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NCERT Exemplar · Class 11 Mathematics Permutations and Combinations

64 questions · 64 still being checked

EXERCISE 7.3 51–60 (part 6 of 7)

  1. State whether the statements in Exercises from $\displaystyle 51$ to $\displaystyle 59$ True or False? Also give justification.

    Exercise 51

    There are 12\displaystyle 12 points in a plane of which 5\displaystyle 5 points are collinear, then the number of lines obtained by joining these points in pairs is 12C2−5C2\displaystyle { }^{12} \mathrm{C}_2-{ }^5 \mathrm{C}_2.

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    NCERT’s answer
    False
    FalseThe $\displaystyle 5$ collinear points give only one line, yet \(\displaystyle {}^{5}\mathrm{C}_2\) removes all $\displaystyle 10$ pairs without adding that one line back.\[{}^{12}\mathrm{C}_2 - {}^{5}\mathrm{C}_2 + 1 = 66 - 10 + 1 = 57 \ne 56 \]

  2. Exercise 52

    Three letters can be posted in five letterboxes in 35\displaystyle 3^5 ways.

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    NCERT’s answer
    False
    FalseEach letter independently goes into any one of the $\displaystyle 5$ boxes.\[5\times 5\times 5 = 5^3 = 125 \ne 3^5 \]

  3. Exercise 53

    In the permutations of n\displaystyle n things, r\displaystyle r taken together, the number of permutations in which m\displaystyle m particular things occur together is n−mPr−m×rPm\displaystyle { }^{n-m} \mathrm{P}_{r-m} \times{ }^r \mathrm{P}_m.

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    NCERT’s answer
    False
    FalseTogether means side by side: the \(\displaystyle m\) things form one block. The stated product counts only arrangements where they are all included.\[{}^{n-m}\mathrm{P}_{r-m}\times{}^{r}\mathrm{P}_m = {}^{n-m}\mathrm{C}_{r-m}\times r! \]Block count:\[{}^{n-m}\mathrm{C}_{r-m}\times (r-m+1)!\times m! \]Check, \(\displaystyle n=4,\ r=3,\ m=2\):\[{}^{2}\mathrm{C}_1\times 2!\times 2! = 8 \]\[{}^{2}\mathrm{P}_1\times{}^{3}\mathrm{P}_2 = 2\times 6 = 12 \ne 8 \]

  4. Exercise 54

    In a steamer there are stalls for 12\displaystyle 12 animals, and there are horses, cows and calves (not less than 12\displaystyle 12 each) ready to be shipped. They can be loaded in 312\displaystyle 3^{12} ways.

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    NCERT’s answer
    True
    TrueEach of the $\displaystyle 12$ stalls takes a horse, a cow or a calf, independently; each kind has at least $\displaystyle 12$ available.\[\underbrace{3\times 3\times\cdots\times 3}_{12} = 3^{12} \]

  5. Exercise 55

    If some or all of n\displaystyle n objects are taken at a time, the number of combinations is 2n−1\displaystyle 2^n-1.

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    NCERT’s answer
    True
    TrueSome or all of \(\displaystyle n\) distinct objects means at least one is taken.\[{}^{n}\mathrm{C}_1 + {}^{n}\mathrm{C}_2 + \cdots + {}^{n}\mathrm{C}_n = 2^n - {}^{n}\mathrm{C}_0 = 2^n - 1 \]

  6. Exercise 56

    There will be only 24\displaystyle 24 selections containing at least one red ball out of a bag containing 4\displaystyle 4 red and 5\displaystyle 5 black balls. It is being given that the balls of the same colour are identical.

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    NCERT’s answer
    True
    TrueIdentical balls: only how many of each colour matters. Red: $\displaystyle 1$ to $\displaystyle 4$, so $\displaystyle 4$ ways. Black: $\displaystyle 0$ to $\displaystyle 5$, so $\displaystyle 6$ ways.\[4\times 6 = 24 \]\[(4+1)(5+1) - 1\cdot(5+1) = 30 - 6 = 24 \]

  7. Exercise 57

    Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table. The number of ways in which the seating arrangements can be made is 11!5!6!(9!)(9!)\displaystyle \frac{11!}{5!6!}(9!)(9!). [Hint: After sending 4\displaystyle 4 on one side and 3\displaystyle 3 on the other side, we have to select out of 11\displaystyle 11; 5\displaystyle 5 on one side and 6\displaystyle 6 on the other. Now there are 9\displaystyle 9 on each side of the long table and each can be arranged in 9\displaystyle 9! ways.]

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    NCERT’s answer
    True
    True. Seat the $\displaystyle 4$ and $\displaystyle 3$ fixed guests first; the other $\displaystyle 11$ guests fill $\displaystyle 5$ more seats on one side and $\displaystyle 6$ on the other. \[{}^{11}\mathrm{C}_5 = \frac{11!}{5!\,6!} \] Each side now holds $\displaystyle 9$ guests, arranged in \(\displaystyle 9!\) ways. \[\text{Total} = \frac{11!}{5!\,6!}\times 9!\times 9! \]

  8. Exercise 58

    A candidate is required to answer 7\displaystyle 7 questions out of 12\displaystyle 12 questions which are divided into two groups, each containing 6\displaystyle 6 questions. He is not permitted to attempt more than 5\displaystyle 5 questions from either group. He can choose the seven questions in 650\displaystyle 650 ways.

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    NCERT’s answer
    False
    False. At most $\displaystyle 5$ from each group, so the $\displaystyle 7$ questions split \(\displaystyle 2+5,\ 3+4,\ 4+3,\ 5+2\). \[2\left[{}^{6}\mathrm{C}_2\,{}^{6}\mathrm{C}_5+{}^{6}\mathrm{C}_3\,{}^{6}\mathrm{C}_4\right] = 2\,(15\cdot 6+20\cdot 15) \] \[= 2\times 390 = 780 \neq 650 \]

  9. Exercise 59

    To fill 12\displaystyle 12 vacancies there are 25\displaystyle 25 candidates of which 5\displaystyle 5 are from scheduled castes. If 3\displaystyle 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is 5C3×20C9\displaystyle { }^5 \mathrm{C}_3 \times{ }^{20} \mathrm{C}_9.

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    NCERT’s answer
    False
    False. Pick the $\displaystyle 3$ scheduled caste candidates for the reserved vacancies; the $\displaystyle 9$ open vacancies are filled from the other $\displaystyle 22$ candidates, not 20. \[{}^{5}\mathrm{C}_3\times{}^{22}\mathrm{C}_9 = 10\times 497420 = 4974200 \] \[{}^{5}\mathrm{C}_3\times{}^{20}\mathrm{C}_9 = 1679600 \neq 4974200 \]
  10. In each if the Exercises from $\displaystyle 60$ to $\displaystyle 64$ match each item given under the column \(\displaystyle \mathrm{C}_1\) to its correct answer given under the column \(\displaystyle \mathrm{C}_2\).

    Exercise 60

    There are 3\displaystyle 3 books on Mathematics, 4\displaystyle 4 on Physics and 5\displaystyle 5 on English. How many different collections can be made such that each collection consists of :
    C1\displaystyle \mathbf{C}_1C2\displaystyle \mathbf{C}_2
    (a) One book of each subject;(i) 3968\displaystyle 3968
    (b) At least one book of each subject :(ii) 60\displaystyle 60
    (c) At least one book of English:(iii) 3255\displaystyle 3255

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    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (ii)
    (b)
    \(\displaystyle \leftrightarrow\) (iii) and
    (c)
    \(\displaystyle \leftrightarrow\) (i)
    (a)-(ii), (b)-(iii), (c)-(i)
    (a)
    One book from each subject:
    \[3\times 4\times 5 = 60 \]
    (b)
    Each subject gives a non-empty collection:
    \[(2^3-1)(2^4-1)(2^5-1) = 7\times 15\times 31 = 3255 \]
    (c)
    At least one of the $\displaystyle 5$ English books, any choice from the other $\displaystyle 7$ books:
    \[(2^5-1)\times 2^{7} = 31\times 128 = 3968 \]