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NCERT Exemplar · Class 11 Mathematics Permutations and Combinations

64 questions · 64 still being checked

EXERCISE 7.3 21–30 (part 3 of 7)

  1. Exercise 21

    18\displaystyle 18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

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    NCERT’s answer
    \(\displaystyle \frac{18!}{(6!)^3}\)
    The three groups are distinct (two experimental, one control), each of $\displaystyle 6$ mice. \[{}^{18}\mathrm{C}_6 \times {}^{12}\mathrm{C}_6 \times {}^{6}\mathrm{C}_6 \] \[= 18564 \times 924 \times 1 \] \[= \frac{18!}{(6!)^3} = 17\,153\,136 \] Answer: \(\displaystyle \dfrac{18!}{(6!)^3} = 17\,153\,136\) ways

  2. Exercise 22

    A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if
    (a)
    they can be of any colour
    (b)
    two must be white and two red and
    (c)
    they must all be of the same colour.

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    NCERT’s answer
    (a)
    \(\displaystyle 11 \mathrm{C}_4\)
    (b)
    \(\displaystyle 6 \mathrm{C}_2 \times 5 \mathrm{C}_2\)
    (c)
    \(\displaystyle 6 \mathrm{C}_4+5 \mathrm{C}_4\)
    Total marbles \(\displaystyle 6 + 5 = 11\).
    (a)
    Any colour:
    \[{}^{11}\mathrm{C}_4 = 330 \]
    (b)
    $\displaystyle 2$ white and $\displaystyle 2$ red:
    \[{}^{6}\mathrm{C}_2 \times {}^{5}\mathrm{C}_2 = 15 \times 10 = 150 \]
    (c)
    All white or all red:
    \[{}^{6}\mathrm{C}_4 + {}^{5}\mathrm{C}_4 = 15 + 5 = 20 \]
    Answer: (a) $\displaystyle 330$ (b) $\displaystyle 150$ (c) $\displaystyle 20$

  3. Exercise 23

    In how many ways can a football team of 11\displaystyle 11 players be selected from 16\displaystyle 16 players? How many of them will
    (i)
    include 2\displaystyle 2 particular players?
    (ii)
    exclude 2\displaystyle 2 particular players?

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    NCERT’s answer
    (i)
    \(\displaystyle 14 \mathrm{C}_9\)
    (ii)
    \(\displaystyle 14 \mathrm{C}_{11}\)
    Selecting $\displaystyle 11$ from $\displaystyle 16$:
    \[{}^{16}\mathrm{C}_{11} = {}^{16}\mathrm{C}_{5} = \frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12}{5!} = 4368 \]
    (i)
    Two particular players included; choose the other $\displaystyle 9$ from $\displaystyle 14$:
    \[{}^{14}\mathrm{C}_{9} = {}^{14}\mathrm{C}_{5} = 2002 \]
    (ii)
    Two particular players excluded; choose $\displaystyle 11$ from $\displaystyle 14$:
    \[{}^{14}\mathrm{C}_{11} = {}^{14}\mathrm{C}_{3} = 364 \]
    Answer: $\displaystyle 4368$ ways; (i) $\displaystyle 2002$ (ii) $\displaystyle 364$

  4. Exercise 24

    A sports team of 11\displaystyle 11 students is to be constituted, choosing at least 5\displaystyle 5 from Class XI and atleast 5\displaystyle 5 from Class XII. If there are 20\displaystyle 20 students in each of these classes, in how many ways can the team be constituted?

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    NCERT’s answer
    \(\displaystyle 2\left(20 \mathrm{C}_5 \times 20 \mathrm{C}_6\right)\)
    With $\displaystyle 11$ players and at least $\displaystyle 5$ from each class, the split is $\displaystyle 5$ + $\displaystyle 6$ or $\displaystyle 6$ + 5. \[{}^{20}\mathrm{C}_5 \times {}^{20}\mathrm{C}_6 + {}^{20}\mathrm{C}_6 \times {}^{20}\mathrm{C}_5 \] \[= 2 \times 15504 \times 38760 \] \[= 2 \times 600\,935\,040 \] \[= 1\,201\,870\,080 \] Answer: \(\displaystyle 2\,({}^{20}\mathrm{C}_5 \times {}^{20}\mathrm{C}_6) = 1\,201\,870\,080\) ways

  5. Exercise 25

    A group consists of 4\displaystyle 4 girls and 7\displaystyle 7 boys. In how many ways can a team of 5\displaystyle 5 members be selected if the team has
    (i)
    no girls
    (ii)
    at least one boy and one girl
    (iii)
    at least three girls.

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    NCERT’s answer
    (i)
    $\displaystyle 21$,
    (ii)
    $\displaystyle 441$
    (iii)
    $\displaystyle 91$
    (i)
    No girls, so all $\displaystyle 5$ come from the $\displaystyle 7$ boys:
    \[{}^7C_5 = 21 \]
    (ii)
    Total teams, minus all-boys teams, minus all-girls teams (none, as there are only $\displaystyle 4$ girls):
    \[{}^{11}C_5 - {}^7C_5 - {}^4C_5 = 462 - 21 - 0 = 441 \]
    (iii)
    At least three girls means $\displaystyle 3$ girls and $\displaystyle 2$ boys, or $\displaystyle 4$ girls and $\displaystyle 1$ boy:
    \[{}^4C_3\,{}^7C_2 + {}^4C_4\,{}^7C_1 = 4\times 21 + 1\times 7 = 91 \]
    Answer: (i) \(\displaystyle 21\) (ii) \(\displaystyle 441\) (iii) \(\displaystyle 91\)
  6. 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 26

    If nC12=nC8\displaystyle { }^n \mathrm{C}_{12}={ }^n \mathrm{C}_8, then n\displaystyle n is equal to
    (A)
    20\displaystyle 20
    (B)
    12\displaystyle 12
    (C)
    6\displaystyle 6 (D) 30\displaystyle 30

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    NCERT’s answer
    A
    (A) \(\displaystyle 20\). Use \(\displaystyle {}^nC_r={}^nC_{n-r}\): \[{}^nC_{12} = {}^nC_{n-12} = {}^nC_8 \] \[n-12 = 8 \] \[n = 20 \]

  7. Exercise 27

    The number of possible outcomes when a coin is tossed 6\displaystyle 6 times is
    (A)
    36\displaystyle 36
    (B)
    64\displaystyle 64
    (C)
    12\displaystyle 12
    (D)
    32\displaystyle 32

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    B
    (B) \(\displaystyle 64\). Each toss has $\displaystyle 2$ outcomes, independent of the others: \[2^6 = 64 \]

  8. Exercise 28

    The number of different four digit numbers that can be formed with the digits 2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4, 7\displaystyle 7 and using each digit only once is
    (A)
    120\displaystyle 120
    (B)
    96\displaystyle 96
    (C)
    24\displaystyle 24
    (D)
    100\displaystyle 100

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    NCERT’s answer
    C
    (C) \(\displaystyle 24\). All four digits are used once, so the numbers are the arrangements of $\displaystyle 4$ digits: \[4! = 24 \]

  9. Exercise 29

    The sum of the digits in unit place of all the numbers formed with the help of 3\displaystyle 3, 4\displaystyle 4, 5\displaystyle 5 and 6\displaystyle 6 taken all at a time is
    (A)
    432\displaystyle 432
    (B)
    108\displaystyle 108
    (C)
    36\displaystyle 36
    (D)
    18\displaystyle 18

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    NCERT’s answer
    B
    (B) \(\displaystyle 108\). Each digit occupies the units place in as many numbers as the other three digits can be arranged: \[3! = 6 \] \[6\,(3+4+5+6) = 6\times 18 = 108 \]

  10. Exercise 30

    Total number of words formed by 2\displaystyle 2 vowels and 3\displaystyle 3 consonants taken from 4\displaystyle 4 vowels and 5\displaystyle 5 consonants is equal to
    (A)
    60\displaystyle 60
    (B)
    120\displaystyle 120
    (C)
    7200\displaystyle 7200
    (D)
    720\displaystyle 720

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    NCERT’s answer
    C
    (C) \(\displaystyle 7200\). Select the letters, then arrange all $\displaystyle 5$: \[{}^4C_2 \times {}^5C_3 = 6\times 10 = 60 \] \[60\times 5! = 60\times 120 = 7200 \]