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NCERT Exemplar · Class 11 Mathematics Permutations and Combinations

64 questions · 64 still being checked

EXERCISE 7.3 41–50 (part 5 of 7)

  1. Fill in the Blanks in the Exercises $\displaystyle 41$ to 50.

    Exercise 41

    If nPr=840,nCr=35\displaystyle { }^n \mathrm{P}_r=840,{ }^n \mathrm{C}_r=35, then r=\displaystyle r= ____\displaystyle \_\_\_\_.

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    \(\displaystyle r = 4\)\[{}^n\mathrm{P}_r = r!\,\cdot {}^n\mathrm{C}_r \]\[r! = \frac{840}{35} = 24 = 4! \]\[r = 4 \]Check: \(\displaystyle n(n-1)(n-2)(n-3) = 840\) gives \(\displaystyle n = 7\), and\[{}^7\mathrm{C}_4 = \frac{7\cdot 6\cdot 5}{3!} = 35 \]

  2. Exercise 42

    15C8+15C9−15C6−15C7=\displaystyle { }^{15} \mathrm{C}_8+{ }^{15} \mathrm{C}_9-{ }^{15} \mathrm{C}_6-{ }^{15} \mathrm{C}_7= ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 0$
    \(\displaystyle 0\)\[{}^{n}\mathrm{C}_r = {}^{n}\mathrm{C}_{n-r} \Rightarrow {}^{15}\mathrm{C}_8 = {}^{15}\mathrm{C}_7, \quad {}^{15}\mathrm{C}_9 = {}^{15}\mathrm{C}_6 \]\[{}^{15}\mathrm{C}_7 + {}^{15}\mathrm{C}_6 - {}^{15}\mathrm{C}_6 - {}^{15}\mathrm{C}_7 = 0 \]

  3. Exercise 43

    The number of permutations of n\displaystyle n different objects, taken r\displaystyle r at a line, when repetitions are allowed, is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle n^r\)
    \(\displaystyle n^r\)Each of the \(\displaystyle r\) places can be filled in \(\displaystyle n\) ways.\[\underbrace{n \times n \times \cdots \times n}_{r \text{ times}} = n^r \]

  4. Exercise 44

    The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is ____\displaystyle \_\_\_\_. [Hint: Number of ways of arranging 6\displaystyle 6 consonants of which two are alike is 6!2!\displaystyle \frac{6!}{2!} and number of ways of arranging vowels =7P6×13!×12!\displaystyle ={ }^7 \mathrm{P}_6 \times \frac{1}{3!} \times \frac{1}{2!}.]

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    NCERT’s answer
    $\displaystyle 1,51,200$
    \(\displaystyle 151200\)Consonants: N, T, R, M, D, T (T twice). Vowels: I, E, E, I, A, E (I twice, E three times).Arrange the consonants:\[\frac{6!}{2!} = 360 \]They leave $\displaystyle 7$ gaps; the $\displaystyle 6$ vowels go one to a gap, so none are together:\[\frac{{}^7\mathrm{P}_6}{3!\,2!} = \frac{5040}{12} = 420 \]\[360 \times 420 = 151200 \]

  5. Exercise 45

    Three balls are drawn from a bag containing 5\displaystyle 5 red, 4\displaystyle 4 white and 3\displaystyle 3 black balls. The number of ways in which this can be done if at least 2\displaystyle 2 are red is ____\displaystyle \_\_\_\_

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    NCERT’s answer
    $\displaystyle 80$
    \(\displaystyle 80\)At least $\displaystyle 2$ red means $\displaystyle 2$ red and $\displaystyle 1$ non-red ($\displaystyle 4$ white + $\displaystyle 3$ black = $\displaystyle 7$), or $\displaystyle 3$ red.\[{}^5\mathrm{C}_2 \times {}^7\mathrm{C}_1 + {}^5\mathrm{C}_3 \]\[= 10 \times 7 + 10 = 80 \]

  6. Exercise 46

    The number of six-digit numbers, all digits of which are odd is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle 5^6\)
    \(\displaystyle 15625\)Each of the $\displaystyle 6$ places can hold any of the odd digits \(\displaystyle 1, 3, 5, 7, 9\).\[5^6 = 15625 \]

  7. Exercise 47

    In a football championship, 153\displaystyle 153 matches were played . Every two teams played one match with each other. The number of teams, participating in the championship is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 18$
    \(\displaystyle 18\)Each match is a choice of $\displaystyle 2$ teams from \(\displaystyle n\).\[{}^n\mathrm{C}_2 = 153 \]\[n(n-1) = 306 \]\[n^2 - n - 306 = 0 \]\[(n-18)(n+17) = 0 \]Since \(\displaystyle n > 0\), \(\displaystyle n = 18\).

  8. Exercise 48

    The total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two signs '-' occur together is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 35$
    \(\displaystyle 35\)Place the six \(\displaystyle +\) signs first. They leave $\displaystyle 7$ gaps, counting both ends. Put each \(\displaystyle -\) in a different gap so no two are together.\[{}^7\mathrm{C}_4 = \frac{7\cdot 6\cdot 5\cdot 4}{4!} = 35 \]

  9. Exercise 49

    A committee of 6\displaystyle 6 is to be chosen from 10\displaystyle 10 men and 7\displaystyle 7 women so as to contain atleast 3\displaystyle 3 men and 2\displaystyle 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee. [Hint:At least 3\displaystyle 3 men and 2\displaystyle 2 women: The number of ways =10C3×7C3+10C4×7C2\displaystyle ={ }^{10} \mathrm{C}_3 \times{ }^7 \mathrm{C}_3+{ }^{10} \mathrm{C}_4 \times{ }^7 \mathrm{C}_2. For 2\displaystyle 2 particular women to be always there: the number of ways =10C4+10C3×5C1\displaystyle ={ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_3 \times{ }^5 \mathrm{C}_1. The total number of committees when two particular women are never together = Total - together.]

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    NCERT’s answer
    $\displaystyle 7800$
    $\displaystyle 7800$At least $\displaystyle 3$ men and $\displaystyle 2$ women in a committee of $\displaystyle 6$ gives $\displaystyle 3$ men + $\displaystyle 3$ women or $\displaystyle 4$ men + $\displaystyle 2$ women.\[\text{Total} = {}^{10}\mathrm{C}_3\times{}^{7}\mathrm{C}_3 + {}^{10}\mathrm{C}_4\times{}^{7}\mathrm{C}_2 = 120\times 35 + 210\times 21 = 8610 \]Both particular women serve ($\displaystyle 3$ men + $\displaystyle 1$ other woman, or $\displaystyle 4$ men):\[\text{Together} = {}^{10}\mathrm{C}_3\times{}^{5}\mathrm{C}_1 + {}^{10}\mathrm{C}_4 = 600 + 210 = 810 \]\[8610 - 810 = 7800 \]

  10. Exercise 50

    A box contains 2\displaystyle 2 white balls, 3\displaystyle 3 black balls and 4\displaystyle 4 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 64$
    $\displaystyle 64$Nine balls in all; a draw with no black ball uses only the $\displaystyle 6$ others.\[{}^{9}\mathrm{C}_3 - {}^{6}\mathrm{C}_3 = 84 - 20 = 64 \]