Question 1

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?

Accepted Answer

(i) 125, (ii) 60.. Fundamental principle of counting. A 3-digit number has three places — hundreds, tens and units — and each place must be filled by one of the digits 1, 2, 3, 4, 5. None of these digits is 0, so every filling gives a genuine 3-digit number. (i) Repetition allowed. The choice made at one place puts no restriction on the others, so each of the three places can be filled in 5 ways. 5 × 5 × 5 = 125 (ii) Repetition not allowed. Fill the hundreds place in 5 ways. One digit is now used up, so the tens place can be filled in 4 ways, and the units place in 3 ways. 5 × 4 × 3 = 60 This is just ^5P 3. (i) 125 numbers, (ii) 60 numbers.

Question 2

How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Accepted Answer

108. Fill the restricted place first. A number is even exactly when its units digit is even, so the units place is the one carrying a condition — settle it before the free places. Units place: the even digits available are 2, 4, 6, so it can be filled in 3 ways. Repetition is allowed, so the digit used at the units place is still available elsewhere. The tens place can be filled in 6 ways and the hundreds place in 6 ways (no digit is 0, so the hundreds place needs no separate care). By the fundamental principle of counting, 6 × 6 × 3 = 108 108 such 3-digit even numbers can be formed.

Question 3

How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Accepted Answer

5040. Fundamental principle of counting. The first 10 letters of the alphabet are a, b, c, …, j. A code is an ordered string of 4 letters, and no letter may repeat. Fill the four positions one after another. The first can be any of the 10 letters; the second any of the 9 still unused; the third any of the remaining 8; the fourth any of the remaining 7. 10 × 9 × 8 × 7 = 5040 Equivalently ^10P 4 = (10!)/(6!) = 5040. 5040 four-letter codes can be formed.

Question 4

How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Accepted Answer

336. Fix the given digits, then count the rest. The number has 5 places, and the first two are already decided: they must be 6 and 7. 6 7 x x x Since no digit may appear more than once, the digits 6 and 7 are now unavailable, leaving 10 - 2 = 8 digits for the last three places. Third place: 8 ways. Fourth place: 7 ways. Fifth place: 6 ways. 8 × 7 × 6 = 336 That is ^8P 3 = 336. 336 such telephone numbers can be constructed.

Question 5

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Accepted Answer

8. Multiplication principle. Each toss is recorded separately, and each toss has 2 possible outcomes — head ( H ) or tail ( T ). The result of one toss does not restrict the next, so the three tosses can be combined freely. 2 × 2 × 2 = 2^3 = 8 The outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. There are 8 possible outcomes.

Question 6

Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Accepted Answer

20. Ordered selection. A signal uses 2 of the 5 flags, one placed below the other. The two positions — upper and lower — are distinguishable, so a red flag above a blue flag is a different signal from a blue flag above a red one. Order matters, so this is a permutation. Upper position: any of the 5 flags, so 5 ways. Lower position: that flag is used, so 4 ways. 5 × 4 = 20 That is ^5P 2 = (5!)/(3!) = 20. 20 different signals can be generated.

Question 7

Evaluate (i) 8! (ii) 4! - 3!

Accepted Answer

(i) 40320, (ii) 18. Direct expansion. By definition n! = n(n-1)(n-2) 3 · 2 · 1. (i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320 (ii) 4! = 24 and 3! = 6, so 4! - 3! = 24 - 6 = 18 A neater route is to take out the common factor: 4! - 3! = 4 · 3! - 3! = 3!(4-1) = 6 × 3 = 18. (i) 8! = 40320 (ii) 4! - 3! = 18.

Question 8

Is 3!+4!=7!?

Accepted Answer

30, No. Evaluate both sides. Never assume the factorial splits over a sum — check by computing. Left side: 3! + 4! = 6 + 24 = 30 Right side: 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 Since 30 ≠ 5040, the two sides are not equal. The tempting step 3! + 4! = (3+4)! is wrong: the factorial is defined by a product, so it does not distribute over addition. No — 3! + 4! = 30 while 7! = 5040, so 3! + 4! ≠ 7!.

Question 9

Compute (8!)/(6!× 2!)

Accepted Answer

28. Cancel the common factorial. Rather than multiplying 8! out, write the larger factorial in terms of the smaller one: 8! = 8 × 7 × 6! Therefore (8!)/(6! × 2!) = (8 × 7 × 6!)/(6! × 2!) = (8 × 7)/(2 × 1) = 56/2 = 28 (8!)/(6! × 2!) = 28.

Question 10

If 1/(6!)+1/(7!)=x/(8!), find x

Accepted Answer

64. Clear the factorials. Multiply both sides of 1/(6!) + 1/(7!) = x/(8!) by 8!, the largest factorial present. This turns every term into a whole number. (8!)/(6!) + (8!)/(7!) = x Now use 8! = 8 × 7 × 6! and 8! = 8 × 7!: (8!)/(6!) = 8 × 7 = 56, (8!)/(7!) = 8 Hence x = 56 + 8 = 64 x = 64.

Question 11

Evaluate (n!)/((n-r)!), when (i) n=6, r=2 (ii) n=9, r=5.

Accepted Answer

(i) 30, (ii) 15120. Cancel the common factorial. In (n!)/((n-r)!) the denominator is a factor of the numerator, so the quotient is simply the product of the top r factors of n!. (i) n = 6, r = 2. (6!)/((6-2)!) = (6!)/(4!) = (6 × 5 × 4!)/(4!) = 6 × 5 = 30 (ii) n = 9, r = 5. (9!)/((9-5)!) = (9!)/(4!) = (9 × 8 × 7 × 6 × 5 × 4!)/(4!) = 9 × 8 × 7 × 6 × 5 = 15120 (i) 30 (ii) 15120.

Question 12

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Accepted Answer

504. Fundamental principle of counting. The available digits are 1, 2, 3, …, 9 — nine digits, none of them 0, so any arrangement of three of them is a valid 3-digit number. No digit may repeat, so fill the places in turn: - hundreds place: 9 ways, - tens place: 8 ways (one digit already used), - units place: 7 ways. 9 × 8 × 7 = 504 This is the permutation ^9P 3 = (9!)/(6!) = 504. 504 such 3-digit numbers can be formed.

Question 13

How many 4-digit numbers are there with no digit repeated?

Accepted Answer

4536. Fill the leading place first. All ten digits 0, 1, 2, …, 9 are available, but a 4-digit number cannot begin with 0 — otherwise it would be a 3-digit number. So the thousands place is the restricted one and must be settled first. Thousands place: any of 1, 2, …, 9, so 9 ways. One digit is now used, leaving 9 digits (including 0 ) for the other three places, with no repetition: 9 × 8 × 7 = 504 By the fundamental principle of counting, 9 × 504 = 4536 There are 4536 four-digit numbers with no digit repeated.

Question 14

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Accepted Answer

60. Fill the restricted place first. The digits available are 1, 2, 3, 4, 6, 7 — six digits, none of them 0. A number is even precisely when its units digit is even, so the units place carries the condition. Units place: the even digits in the list are 2, 4, 6, giving 3 ways. No digit may repeat, so after the units digit is chosen, 5 digits remain: - tens place: 5 ways, - hundreds place: 4 ways. 3 × 5 × 4 = 60 60 such 3-digit even numbers can be made.

Question 15

Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Accepted Answer

120, 48. Permutation, then the restricted place. Five digits 1, 2, 3, 4, 5 are available and none may repeat, so a 4-digit number is an arrangement of 4 of the 5 digits. No digit is 0, so there is no leading-digit restriction. Total count. ^5P 4 = (5!)/((5-4)!) = 120/1 = 120 (directly: 5 × 4 × 3 × 2 = 120 ). How many are even. Fill the units place first: it must be 2 or 4, so 2 ways. Four digits then remain for the other three places: 4 × 3 × 2 = 24 Hence the number of even ones is 2 × 24 = 48 120 four-digit numbers in all, of which 48 are even.

Question 16

From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?

Accepted Answer

56. Ordered selection — a permutation. The two posts are different: choosing person A as chairman and B as vice chairman is not the same appointment as B as chairman and A as vice chairman. So order matters, and one person cannot fill both posts. Chairman: any of the 8 persons, so 8 ways. Vice chairman: any of the remaining 7, so 7 ways. 8 × 7 = 56 That is ^8P 2 = (8!)/(6!) = 56. The two posts can be filled in 56 ways.

Question 17

Find n if ^n-1 P 3:^n P 4=1: 9.

Accepted Answer

9. Simplify the ratio before solving. Use ^nP r = (n!)/((n-r)!): ^n-1P 3 = ((n-1)!)/((n-1-3)!) = ((n-1)!)/((n-4)!),^nP 4 = (n!)/((n-4)!) Both have the same denominator (n-4)!, so it cancels in the ratio: (^n-1P 3)/(^nP 4) = ((n-1)!)/(n!) = ((n-1)!)/(n · (n-1)!) = 1/n The given condition is therefore 1/n = 1/9 ⇒ n = 9 This is admissible, since ^n-1P 3 needs n - 1 ≥ 3 and ^nP 4 needs n ≥ 4. Check: ^8P 3 = 8 × 7 × 6 = 336 and ^9P 4 = 9 × 8 × 7 × 6 = 3024, and 336: 3024 = 1: 9. n = 9.

Question 18

Find r if (i) ^5 P r=2^6 P r-1 (ii) ^5 P r=^6 P r-1.

Accepted Answer

(i) 3, (ii) 4. Expand both permutations and cancel. Throughout, ^nP r = (n!)/((n-r)!) with 0 ≤ r ≤ n. Since ^5P r appears on the left, r must satisfy 0 ≤ r ≤ 5 — this is what picks out the acceptable root at the end. (i) ^5P r = 2 ·^6P r-1. (5!)/((5-r)!) = 2 · (6!)/( (6-(r-1) )!) = 2 · (6!)/((7-r)!) Rearranging so the factorials sit together, ((7-r)!)/((5-r)!) = 2 · (6!)/(5!) = 2 × 6 = 12 Now (7-r)! = (7-r)(6-r)(5-r)!, so (5-r)! cancels: (7-r)(6-r) = 12 42 - 13r + r^2 = 12 ⇒ r^2 - 13r + 30 = 0 ⇒ (r-3)(r-10) = 0 So r = 3 or r = 10. But r ≤ 5, so r = 10 is rejected. Check: ^5P 3 = 60 and 2 ·^6P 2 = 2 × 30 = 60. (ii) ^5P r =^6P r-1. The same working without the factor 2: ((7-r)!)/((5-r)!) = (6!)/(5!) = 6 ⇒ (7-r)(6-r) = 6 r^2 - 13r + 42 = 6 ⇒ r^2 - 13r + 36 = 0 ⇒ (r-4)(r-9) = 0 So r = 4 or r = 9, and r = 9 is rejected as r ≤ 5. Check: ^5P 4 = 120 and ^6P 3 = 120. (i) r = 3 (ii) r = 4.

Question 19

How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Accepted Answer

40320. Permutations of distinct letters. The word EQUATION has 8 letters — E, Q, U, A, T, I, O, N — and no letter is repeated. The number of ways of arranging n distinct objects, all taken at a time, is n!. ^8 P 8 = 8! = 8×7×6×5×4×3×2×1 = 40320 40320 words can be formed.

Question 20

How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel?

Accepted Answer

(i) 360, (ii) 720, (iii) 240. Permutations of distinct letters. MONDAY has 6 different letters — M, O, N, D, A, Y — and no letter may be repeated, so filling r places from them can be done in ^6 P r=(6!)/((6-r)!) ways. (i) 4 letters at a time. ^6 P 4=(6!)/(2!)=6×5×4×3=360 (ii) All letters at a time. ^6 P 6=6!=720 (iii) All letters, first letter a vowel. The vowels in MONDAY are O and A, so the first place can be filled in 2 ways. The remaining 5 letters then fill the other 5 places in 5! ways. 2×5! = 2×120 = 240 (i) 360 (ii) 720 (iii) 240

Question 21

In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?

Accepted Answer

Complementary counting with repeated letters. MISSISSIPPI has 11 letters: M once, I four times, S four times, P twice. When n letters contain repeats of p, q, … alike letters, the number of distinct arrangements is (n!)/(p! q! ). Total distinct permutations: (11!)/(4! 4! 2!) = 39916800/(24×24×2) = 34650 Now count the ones in which the four I's do come together. Tie the four I's into a single block. The objects to arrange are that block together with M, S, S, S, S, P, P — that is 8 objects, with S repeated 4 times and P repeated 2 times: (8!)/(4! 2!) = 40320/48 = 840 (The four I's are identical, so the block has only one internal ordering.) Subtract the "together" cases from the total: 34650 - 840 = 33810 In 33810 of the distinct permutations the four I's do not come together.

Question 22

In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?

Accepted Answer

(i) 1814400, (ii) 2419200, (iii) 25401600. Arrangements with a repeated letter, under position constraints. PERMUTATIONS has 12 letters: P, E, R, M, U, T, A, T, I, O, N, S. Only T is repeated (twice); every other letter occurs once. So a divisor of 2! appears in each count. (i) Words start with P and end with S. Those two places are fixed, leaving the 10 letters E, R, M, U, T, A, T, I, O, N for the ten middle places, with T twice: (10!)/(2!) = 3628800/2 = 1814400 (ii) Vowels all together. The vowels are E, U, A, I, O — five distinct letters. Tie them into one block. That block together with the 7 consonants P, R, M, T, T, N, S gives 8 objects to arrange, T repeated twice: (8!)/(2!) = 20160 Inside the block the 5 vowels can be ordered in 5! = 120 ways. 20160×120 = 2419200 (iii) Always 4 letters between P and S. P and S must occupy places whose numbers differ by 5: (1,6), (2,7), (3,8), (4,9), (5,10), (6,11), (7,12) — that is 7 pairs of places, and in each pair P and S can be interchanged, giving 7×2 = 14 choices. The other 10 letters (T twice) fill the remaining ten places in (10!)/(2!) = 1814400 ways. 14×1814400 = 25401600 (i) 1814400 (ii) 2419200 (iii) 25401600

Question 23

If ^n C 8=^n C 2, find ^n C 2.

Accepted Answer

45. Use ^n C a=^n C b⇒ a=b or a+b=n. Here a=8 and b=2 are not equal, so the second alternative must hold: n = 8+2 = 10 Hence ^n C 2=^10 C 2 = (10!)/(2! 8!) = (10×9)/(2×1) = 45 (Check: ^10 C 8=(10×9)/2=45=^10 C 2, as required.) ^n C 2 = 45.

Question 24

Determine n if (i) ^2 n C 3:^n C 3=12: 1 (ii) ^2 n C 3:^n C 3=11: 1

Accepted Answer

(i) 5, (ii) 6. Expand both combinations and cancel. For any m≥ 3, ^m C 3=(m(m-1)(m-2))/(3!). The 3! cancels in the ratio: (^2n C 3)/(^n C 3) = (2n(2n-1)(2n-2))/(n(n-1)(n-2)) = (2n (2n-1) 2(n-1))/(n(n-1)(n-2)) = (4(2n-1))/n-2 (i) Ratio 12:1. (4(2n-1))/n-2=12 ⇒ 8n-4 = 12n-24 ⇒ 4n = 20 ⇒ n = 5 Check: ^10 C 3:^5 C 3 = 120:10 = 12:1. (ii) Ratio 11:1. (4(2n-1))/n-2=11 ⇒ 8n-4 = 11n-22 ⇒ 3n = 18 ⇒ n = 6 Check: ^12 C 3:^6 C 3 = 220:20 = 11:1. (i) n = 5 (ii) n = 6

Question 25

How many chords can be drawn through 21 points on a circle?

Accepted Answer

210. Combinations — a chord is fixed by its two endpoints. A chord is drawn by joining two of the given points, and the order in which the two endpoints are named does not matter. No three points of a circle are collinear, so every distinct pair of points gives a distinct chord. ^21 C 2 = (21!)/(2! 19!) = (21×20)/(2×1) = 210 210 chords can be drawn.

Question 26

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Accepted Answer

40. Choose each group separately, then multiply. The boys are chosen from the boys and the girls from the girls, so the two choices are independent and the multiplication principle applies. Order within a team does not matter, so use combinations. ^5 C 3×^4 C 3 = (5×4×3)/(3×2×1)×(4×3×2)/(3×2×1) = 10×4 = 40 40 ways.

Question 27

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Accepted Answer

2000. One combination per colour, then multiply. Each selection is to consist of 3 balls of each colour, and 3+3+3=9, so the whole selection is fixed by choosing 3 red from 6, 3 white from 5 and 3 blue from 5. The three pools are separate, so multiply. ^6 C 3×^5 C 3×^5 C 3 = (6×5×4)/(3×2×1)×(5×4×3)/(3×2×1)×(5×4×3)/(3×2×1) = 20×10×10 = 2000 2000 ways.

Question 28

Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Accepted Answer

778320. Split the deck into aces and non-aces. A deck of 52 cards contains 4 aces and 52-4=48 other cards. "Exactly one ace" means 1 card is taken from the 4 aces and the remaining 4 cards from the 48 non-aces. ^4 C 1×^48 C 4 = 4×(48×47×46×45)/(4×3×2×1) = 4×194580 = 778320 778320 such 5-card combinations.

Question 29

In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Accepted Answer

3960. Split the squad into bowlers and non-bowlers. Of the 17 players, 5 can bowl, so 17-5=12 cannot. The team of 11 must contain exactly 4 bowlers, so the other 11-4=7 places are filled from the 12 non-bowlers. ^5 C 4×^12 C 7 ^5 C 4=5,^12 C 7=^12 C 5=(12×11×10×9×8)/(5×4×3×2×1)=792 5×792 = 3960 3960 ways.

Question 30

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Accepted Answer

200. Choose from each colour separately, then multiply. The 2 black balls come from the 5 black ones and the 3 red balls from the 6 red ones; the two choices are independent. ^5 C 2×^6 C 3 = (5×4)/(2×1)×(6×5×4)/(3×2×1) = 10×20 = 200 200 ways.

Question 31

In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Accepted Answer

35. Fix the compulsory courses, then choose the rest. Two of the 5 courses are already decided for every student, so the only real choice is the remaining 5-2=3 courses, taken from the 9-2=7 courses that are not compulsory. ^7 C 3 = (7×6×5)/(3×2×1) = 35 35 ways.

Question 32

How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Accepted Answer

3600. Select, then arrange. DAUGHTER has 8 distinct letters: the vowels A, U, E (three of them) and the consonants D, G, H, T, R (five of them). Selecting the letters — 2 vowels from 3 and 3 consonants from 5: ^3 C 2×^5 C 3 = 3×10 = 30 Arranging each such choice — every selection has 5 different letters, so it gives 5! = 120 words. 30×120 = 3600 3600 words.

Question 33

How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?

Accepted Answer

1440. Group-and-arrange method. EQUATION has 8 distinct letters: the vowels A, E, I, O, U (five) and the consonants Q, T, N (three). Requiring the vowels to occur together and the consonants to occur together means all five vowels form one unbroken block and all three consonants form another. The two blocks can be placed in 2! orders (vowel block first, or consonant block first). Inside the blocks, the vowels can be ordered in 5! ways and the consonants in 3! ways. 2!×5!×3! = 2×120×6 = 1440 1440 words.

Question 34

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls? (ii) atleast 3 girls? (iii) atmost 3 girls?

Accepted Answer

(i) 504, (ii) 588, (iii) 1632. Case-split on the number of girls. A committee of 7 is chosen from 9 boys and 4 girls. Note that only 4 girls exist, so a committee can hold at most 4 of them. (i) Exactly 3 girls — 3 girls and 7-3=4 boys: ^4 C 3×^9 C 4 = 4×126 = 504 (ii) At least 3 girls — 3 girls or 4 girls: ^4 C 3×^9 C 4 504 +^4 C 4×^9 C 3 1×84 = 84 = 588 (iii) At most 3 girls — 0, 1, 2 or 3 girls: ^4 C 0×^9 C 7 = 1×36 = 36 ^4 C 1×^9 C 6 = 4×84 = 336 ^4 C 2×^9 C 5 = 6×126 = 756 ^4 C 3×^9 C 4 = 4×126 = 504 36+336+756+504 = 1632 Check: all committees number ^13 C 7=1716, and removing the only excluded case (4 girls, 84 committees) leaves 1716-84=1632. (i) 504 (ii) 588 (iii) 1632

Question 35

If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?

Accepted Answer

907200. Count the whole alphabetical block that precedes the E-words. EXAMINATION has 11 letters: A twice, I twice, N twice, and E, X, M, T, O once each. In dictionary order the distinct first letters, listed increasingly, are A, E, I, M, N, O, T, X. So every word listed before the first word beginning with E is exactly a word beginning with A, and all of them come before it. Fix one A in the first place. The remaining 10 letters are A, E, I, I, M, N, N, O, T, X, in which I occurs twice and N occurs twice: (10!)/(2! 2!) = 3628800/4 = 907200 907200 words appear in the list before the first word starting with E.

Question 36

How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Accepted Answer

120. Fix the forced digit first. A number is divisible by 10 exactly when its units digit is 0. A 6-digit number made from the six digits 0, 1, 3, 5, 7, 9 with no repetition must use each digit once, so the 0 is forced into the units place and can go nowhere else. The remaining five digits 1, 3, 5, 7, 9 then fill the other five places in 5! = 5×4×3×2×1 = 120 ways. Since 0 is already used up at the end, the leading digit is automatically non-zero, so every one of these arrangements really is a 6-digit number. 120 such numbers.

Question 37

The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Accepted Answer

50400. Choose, then arrange. A "word" here is any arrangement of the four chosen letters, meaningful or not. Building one takes three independent decisions, so the answer is the product of the three counts. Step 1 — choose the vowels. Order does not matter while selecting, so this is a combination: ^5C 2=(5!)/(2! 3!)=(5× 4)/2=10 Step 2 — choose the consonants. ^21C 2=(21!)/(2! 19!)=(21× 20)/2=210 Step 3 — arrange the 4 chosen letters. They are all different, so they can be placed in the four positions in 4!=24 ways. By the fundamental principle of counting, 10× 210× 24=50400 50400 words can be formed.

Question 38

In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Accepted Answer

420. Case analysis on the split between the parts. Let the student take a questions from Part I (which has 5) and b from Part II (which has 7), with a+b=8, a≥ 3, b≥ 3. From b=8-a≥ 3 we get a≤ 5, and a≤ 5 anyway since Part I has only 5 questions. So a∈ 3,4,5 — exactly three cases. Selecting questions is a combination, since order of attempt is irrelevant. Part I (a) Part II (b) Number of ways --- --- --- 3 5 ^5C 3×^7C 5=10× 21=210 4 4 ^5C 4×^7C 4=5× 35=175 5 3 ^5C 5×^7C 3=1× 35=35 The three cases are mutually exclusive, so add them: 210+175+35=420 The student can select the questions in 420 ways.

Question 39

Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Accepted Answer

^4 C 1 ×^48 C 4. Split the deck into kings and non-kings. A pack of 52 cards contains 4 kings and 52-4=48 other cards. "Exactly one king" means the hand takes 1 card from the 4 kings and the remaining 4 cards from the 48 non-kings. The king: ^4C 1=4 The other four cards: ^48C 4=(48× 47× 46× 45)/(4× 3× 2× 1)=194580 The two choices are independent, so multiply: 4× 194580=778320 There are 778320 such 5-card combinations.

Question 40

It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

Accepted Answer

2880. Fix the positions first, then permute within them. There are 5+4=9 people, so the row has 9 seats numbered 1 to 9. The even-numbered seats are 2, 4, 6, 8 — exactly 4 of them, which is precisely the number of women. So the women must fill seats 2, 4, 6, 8 and the men are forced into the five odd seats 1, 3, 5, 7, 9. There is no choice about which seats each group gets; only the order within a group varies. Women among the 4 even seats: 4!=24 ways. Men among the 5 odd seats: 5!=120 ways. These are independent, so 4!× 5!=24× 120=2880 2880 arrangements are possible.

Question 41

From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Accepted Answer

^22 C 7+^22 C 10. Two exclusive cases: the trio is in, or the trio is out. Apart from the 3 students who move as a block, there are 25-3=22 other students. Case 1 — all 3 join. They occupy 3 of the 10 places, so the remaining 10-3=7 members come from the other 22: ^22C 7=170544 Case 2 — none of the 3 joins. All 10 members come from the other 22: ^22C 10=646646 No party can satisfy both conditions at once, so add the cases: 170544+646646=817190 The party can be chosen in ^22C 7+^22C 10=817190 ways.

Question 42

In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?

Accepted Answer

151200. Tie the block together, then use the repeated-letters formula. ASSASSINATION has 13 letters: A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and T, O appear once each (3+4+2+2+1+1=13). Since all four S's must stay together, glue them into a single block [ SSSS ]. We are now arranging A, A, A, I, I, N, N, T, O and the block which is 9+1=10 objects in a row. Among these 10 objects, A repeats 3 times, I repeats 2 times and N repeats 2 times. The number of arrangements of n objects of which p 1,p 2,… are alike is (n!)/(p 1! p 2! ), so (10!)/(3! 2! 2!)=3628800/(6× 2× 2)=3628800/24=151200 The four S's are identical, so permuting them inside the block gives nothing new — no extra factor is needed. 151200 arrangements have all the S's together.

Part I \(\displaystyle (a)\)	Part II \(\displaystyle (b)\)	Number of ways
$\displaystyle 3$	$\displaystyle 5$	\(\displaystyle {}^{5}C_{3}\times{}^{7}C_{5}=10\times 21=210\)
$\displaystyle 4$	$\displaystyle 4$	\(\displaystyle {}^{5}C_{4}\times{}^{7}C_{4}=5\times 35=175\)
$\displaystyle 5$	$\displaystyle 3$	\(\displaystyle {}^{5}C_{5}\times{}^{7}C_{3}=1\times 35=35\)

EXERCISE 6.1 1–6 (part 1 of 5)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6