SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Permutations and Combinations

42 questions · 42 still being checked

EXERCISE 6.4 1–9 (part 4 of 5)

  1. Exercise 1

    If nC8=nC2\displaystyle { }^{n} \mathrm{C}_{8}={ }^{n} \mathrm{C}_{2}, find nC2\displaystyle { }^{n} \mathrm{C}_{2}.

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    NCERT’s answer
    $\displaystyle 45$
    Use \(\displaystyle {}^{n}\mathrm{C}_{a}={}^{n}\mathrm{C}_{b}\Rightarrow a=b\ \text{or}\ a+b=n\). Here \(\displaystyle a=8\) and \(\displaystyle b=2\) are not equal, so the second alternative must hold: \[n = 8+2 = 10\]Hence \[{}^{n}\mathrm{C}_{2}={}^{10}\mathrm{C}_{2} = \frac{10!}{2!\,8!} = \frac{10\times9}{2\times1} = 45\](Check: \(\displaystyle {}^{10}\mathrm{C}_{8}=\dfrac{10\times9}{2}=45={}^{10}\mathrm{C}_{2}\), as required.)\(\displaystyle {}^{n}\mathrm{C}_{2} = 45\).

  2. Exercise 2

    Determine n\displaystyle n if
    (i)
    2nC3:nC3=12:1\displaystyle { }^{2 n} \mathrm{C}_{3}:{ }^{n} \mathrm{C}_{3}=12: 1
    (ii)
    2nC3:nC3=11:1\displaystyle { }^{2 n} \mathrm{C}_{3}:{ }^{n} \mathrm{C}_{3}=11: 1

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    NCERT’s answer
    (i)
    $\displaystyle 5$, (ii) $\displaystyle 6$
    Expand both combinations and cancel. For any \(\displaystyle m\ge 3\), \(\displaystyle {}^{m}\mathrm{C}_{3}=\dfrac{m(m-1)(m-2)}{3!}\). The \(\displaystyle 3!\) cancels in the ratio: \[\frac{{}^{2n}\mathrm{C}_{3}}{{}^{n}\mathrm{C}_{3}} = \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = \frac{2n\,(2n-1)\,2(n-1)}{n(n-1)(n-2)} = \frac{4(2n-1)}{n-2}\](i) Ratio \(\displaystyle 12:1\). \[\frac{4(2n-1)}{n-2}=12 \;\Rightarrow\; 8n-4 = 12n-24 \;\Rightarrow\; 4n = 20 \;\Rightarrow\; n = 5\] Check: \(\displaystyle {}^{10}\mathrm{C}_{3}:{}^{5}\mathrm{C}_{3} = 120:10 = 12:1\).(ii) Ratio \(\displaystyle 11:1\). \[\frac{4(2n-1)}{n-2}=11 \;\Rightarrow\; 8n-4 = 11n-22 \;\Rightarrow\; 3n = 18 \;\Rightarrow\; n = 6\] Check: \(\displaystyle {}^{12}\mathrm{C}_{3}:{}^{6}\mathrm{C}_{3} = 220:20 = 11:1\).(i) \(\displaystyle n = 5\) (ii) \(\displaystyle n = 6\)

  3. Exercise 3

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

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    NCERT’s answer
    $\displaystyle 210$
    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.\[{}^{21}\mathrm{C}_{2} = \frac{21!}{2!\,19!} = \frac{21\times20}{2\times1} = 210\]$\displaystyle 210$ chords can be drawn.

  4. Exercise 4

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

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    NCERT’s answer
    $\displaystyle 40$
    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.\[{}^{5}\mathrm{C}_{3}\times{}^{4}\mathrm{C}_{3} = \frac{5\times4\times3}{3\times2\times1}\times\frac{4\times3\times2}{3\times2\times1} = 10\times4 = 40\]$\displaystyle 40$ ways.

  5. Exercise 5

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

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    NCERT’s answer
    $\displaystyle 2000$
    One combination per colour, then multiply. Each selection is to consist of $\displaystyle 3$ balls of each colour, and \(\displaystyle 3+3+3=9\), so the whole selection is fixed by choosing $\displaystyle 3$ red from $\displaystyle 6$, $\displaystyle 3$ white from $\displaystyle 5$ and $\displaystyle 3$ blue from 5. The three pools are separate, so multiply.\[{}^{6}\mathrm{C}_{3}\times{}^{5}\mathrm{C}_{3}\times{}^{5}\mathrm{C}_{3} = \frac{6\times5\times4}{3\times2\times1}\times\frac{5\times4\times3}{3\times2\times1}\times\frac{5\times4\times3}{3\times2\times1} = 20\times10\times10 = 2000\]$\displaystyle 2000$ ways.

  6. Exercise 6

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

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    NCERT’s answer
    $\displaystyle 778320$
    Split the deck into aces and non-aces. A deck of $\displaystyle 52$ cards contains \(\displaystyle 4\) aces and \(\displaystyle 52-4=48\) other cards. "Exactly one ace" means \(\displaystyle 1\) card is taken from the $\displaystyle 4$ aces and the remaining \(\displaystyle 4\) cards from the $\displaystyle 48$ non-aces.\[{}^{4}\mathrm{C}_{1}\times{}^{48}\mathrm{C}_{4} = 4\times\frac{48\times47\times46\times45}{4\times3\times2\times1} = 4\times194580 = 778320\]$\displaystyle 778320$ such $\displaystyle 5$-card combinations.

  7. Exercise 7

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

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    NCERT’s answer
    $\displaystyle 3960$
    Split the squad into bowlers and non-bowlers. Of the \(\displaystyle 17\) players, \(\displaystyle 5\) can bowl, so \(\displaystyle 17-5=12\) cannot. The team of \(\displaystyle 11\) must contain exactly \(\displaystyle 4\) bowlers, so the other \(\displaystyle 11-4=7\) places are filled from the $\displaystyle 12$ non-bowlers.\[{}^{5}\mathrm{C}_{4}\times{}^{12}\mathrm{C}_{7}\] \[{}^{5}\mathrm{C}_{4}=5,\qquad {}^{12}\mathrm{C}_{7}={}^{12}\mathrm{C}_{5}=\frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}=792\] \[5\times792 = 3960\]$\displaystyle 3960$ ways.

  8. Exercise 8

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

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    NCERT’s answer
    $\displaystyle 200$
    Choose from each colour separately, then multiply. The $\displaystyle 2$ black balls come from the $\displaystyle 5$ black ones and the $\displaystyle 3$ red balls from the $\displaystyle 6$ red ones; the two choices are independent.\[{}^{5}\mathrm{C}_{2}\times{}^{6}\mathrm{C}_{3} = \frac{5\times4}{2\times1}\times\frac{6\times5\times4}{3\times2\times1} = 10\times20 = 200\]$\displaystyle 200$ ways.

  9. Exercise 9

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

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    NCERT’s answer
    $\displaystyle 35$
    Fix the compulsory courses, then choose the rest. Two of the \(\displaystyle 5\) courses are already decided for every student, so the only real choice is the remaining \(\displaystyle 5-2=3\) courses, taken from the \(\displaystyle 9-2=7\) courses that are not compulsory.\[{}^{7}\mathrm{C}_{3} = \frac{7\times6\times5}{3\times2\times1} = 35\]$\displaystyle 35$ ways.