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NCERT Exemplar · Class 11 Mathematics Permutations and Combinations

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EXERCISE 7.3 31–40 (part 4 of 7)

  1. Choose the correct answer out of the given four options against each of the Exercises from $\displaystyle 26$ to $\displaystyle 40$ (M.C.Q.).

    Exercise 31

    A five digit number divisible by 3\displaystyle 3 is to be formed using the numbers 0\displaystyle 0, 1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4 and 5\displaystyle 5 without repetitions. The total number of ways this can be done is
    (A)
    216\displaystyle 216
    (B)
    600\displaystyle 600
    (C)
    240\displaystyle 240
    (D)
    3125\displaystyle 3125
    [Hint: 5\displaystyle 5 digit numbers can be formed using digits 0,1,2,4,5\displaystyle 0,1,2,4,5 or by using digits 1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4, 5\displaystyle 5 since sum of digits in these cases is divisible by 3.]

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    NCERT’s answer
    A
    (A) \(\displaystyle 216\). The digits sum to \[0+1+2+3+4+5 = 15 \] Five digits are used, so one digit \(\displaystyle d\) is left out. The digit sum \(\displaystyle 15-d\) must be divisible by $\displaystyle 3$, so \(\displaystyle d=0\) or \(\displaystyle d=3\). Digits \(\displaystyle 1,2,3,4,5\): \[5! = 120 \] Digits \(\displaystyle 0,1,2,4,5\), with $\displaystyle 0$ not in the first place: \[5! - 4! = 96 \] \[120 + 96 = 216 \]

  2. Exercise 32

    Every body in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
    (A)
    11\displaystyle 11
    (B)
    12\displaystyle 12
    (C)
    13\displaystyle 13
    (D)
    14\displaystyle 14

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    NCERT’s answer
    B
    (B) \(\displaystyle 12\). Each handshake is a pair of persons: \[{}^nC_2 = 66 \] \[\frac{n(n-1)}{2} = 66 \] \[n^2 - n - 132 = 0 \] \[(n-12)(n+11) = 0 \] \[n = 12 \]

  3. Exercise 33

    The number of triangles that are formed by choosing the vertices from a set of 12\displaystyle 12 points, seven of which lie on the same line is
    (A)
    105\displaystyle 105
    (B)
    15\displaystyle 15
    (C)
    175\displaystyle 175
    (D)
    185\displaystyle 185

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    NCERT’s answer
    D
    (D) \(\displaystyle 185\)Any $\displaystyle 3$ of the $\displaystyle 12$ points make a triangle, except $\displaystyle 3$ from the $\displaystyle 7$ collinear points.\[\text{triangles} = {}^{12}C_3 - {}^{7}C_3 \] \[= 220 - 35 \] \[= 185 \]

  4. Exercise 34

    The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is
    (A)
    6\displaystyle 6 (B) 18\displaystyle 18
    (C)
    12\displaystyle 12

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    NCERT’s answer
    B
    (B) \(\displaystyle 18\)A parallelogram needs $\displaystyle 2$ of the $\displaystyle 4$ parallel lines and $\displaystyle 2$ of the $\displaystyle 3$ parallel lines.\[{}^{4}C_2 \times {}^{3}C_2 = 6 \times 3 \] \[= 18 \]

  5. Exercise 35

    The number of ways in which a team of eleven players can be selected from 22\displaystyle 22 players always including 2\displaystyle 2 of them and excluding 4\displaystyle 4 of them is
    (A)
    16C11\displaystyle { }^{16} \mathrm{C}_{11}
    (B)
    16C5\displaystyle { }^{16} \mathrm{C}_5
    (C)
    16C9\displaystyle { }^{16} \mathrm{C}_9
    (D)
    20C9\displaystyle { }^{20} \mathrm{C}_9

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    NCERT’s answer
    C
    (C) \(\displaystyle {}^{16}C_9\)Two players are fixed in, four are out, so $\displaystyle 9$ places remain to be filled from the other players.\[22 - 2 - 4 = 16 \quad \text{players left} \] \[11 - 2 = 9 \quad \text{places left} \] \[\text{ways} = {}^{16}C_9 \]

  6. Exercise 36

    The number of 5\displaystyle 5-digit telephone numbers having atleast one of their digits repeated is
    (A)
    90,000\displaystyle 90,000
    (B)
    10,000\displaystyle 10,000
    (C)
    30,240\displaystyle 30,240
    (D)
    69,760\displaystyle 69,760

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    NCERT’s answer
    D
    (D) \(\displaystyle 69{,}760\)Count all numbers, then remove those with all digits different (a telephone number may begin with $\displaystyle 0$).\[\text{all} = 10^5 = 100000 \] \[\text{all different} = 10 \times 9 \times 8 \times 7 \times 6 = 30240 \] \[\text{required} = 100000 - 30240 = 69760 \]

  7. Exercise 37

    The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is
    (A)
    94\displaystyle 94
    (B)
    126\displaystyle 126
    (C)
    128\displaystyle 128
    (D)
    None

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    NCERT’s answer
    A
    (A) \(\displaystyle 94\)With \(\displaystyle m\) men the committee has \(\displaystyle 2m\) women, \(\displaystyle m \ge 2\). Only \(\displaystyle m = 2, 3\) are possible, since there are $\displaystyle 4$ men and $\displaystyle 6$ women (so \(\displaystyle m=4\) would need $\displaystyle 8$ women).\[m = 2,\ w = 4: \quad {}^{4}C_2 \times {}^{6}C_4 = 6 \times 15 = 90 \] \[m = 3,\ w = 6: \quad {}^{4}C_3 \times {}^{6}C_6 = 4 \times 1 = 4 \] \[\text{total} = 90 + 4 = 94 \]

  8. Exercise 38

    The total number of 9\displaystyle 9 digit numbers which have all different digits is
    (A)
    10\displaystyle 10!
    (B)
    9\displaystyle 9 ! (C) 9×9!\displaystyle 9 \times 9!
    (D)
    10×10!\displaystyle 10 \times 10!

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    NCERT’s answer
    C
    (C) \(\displaystyle 9 \times 9!\)The first digit cannot be 0. Each later digit is chosen from the digits not yet used, $\displaystyle 0$ included.\[\text{first digit} = 9 \ \text{ways} \] \[\text{remaining 8 places} = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 9! \] \[\text{total} = 9 \times 9! \]

  9. Exercise 39

    The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is
    (A)
    1440\displaystyle 1440
    (B)
    144\displaystyle 144
    (C)
    7\displaystyle 7! (D) 4C4×3C3\displaystyle { }^4 \mathrm{C}_4 \times{ }^3 \mathrm{C}_3

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    NCERT’s answer
    B
    (B) \(\displaystyle 144\)ARTICLE has $\displaystyle 3$ vowels (A, I, E) and $\displaystyle 4$ consonants (R, T, C, L). The even places among $\displaystyle 7$ are $\displaystyle 2$, $\displaystyle 4$, 6.\[\text{vowels in 3 even places} = 3! = 6 \] \[\text{consonants in other 4 places} = 4! = 24 \] \[\text{total} = 6 \times 24 = 144 \]

  10. Exercise 40

    Given 5\displaystyle 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
    (A)
    3600\displaystyle 3600
    (B)
    3720\displaystyle 3720
    (C)
    3800\displaystyle 3800
    (D)
    3600\displaystyle 3600
    [Hint: Possible numbers of choosing or not choosing 5\displaystyle 5 green dyes, 4\displaystyle 4 blue dyes and 3\displaystyle 3 red dyes are 25,24\displaystyle 2^5, 2^4 and 23\displaystyle 2^3, respectively.]

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    B
    (B) \(\displaystyle 3720\)Each dye is chosen or not chosen, so a colour of \(\displaystyle n\) dyes gives \(\displaystyle 2^n\) selections; at least one green and one blue removes the empty selection for those two colours.\[\text{green} = 2^5 - 1 = 31 \] \[\text{blue} = 2^4 - 1 = 15 \] \[\text{red} = 2^3 = 8 \] \[\text{total} = 31 \times 15 \times 8 = 3720 \]