Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 6 Ex 6.4 — all 9 questions solved, covering combinations: solving nCr equations, and selecting committees, teams, coloured balls and card hands with fixed conditions.
Every question here is a selection, not an arrangement — {}^{n}C_{r}=\dfrac{n!}{r!(n-r)!} is the only formula in play, since none of these problems care about order. Questions 1 and 2 are algebraic, solved by simplifying the ratio of two combinations. From Question 3 onward, the skill is splitting a compound selection (a fixed number of each colour, a fixed number of bowlers, a compulsory subset) into independent parts, counting each part with nCr, and multiplying the results together.
Using the identity {}^{n}C_{a}={}^{n}C_{b}\ \Rightarrow\ a=b \text{ or } a+b=n, and since 8\ne2:
8+2=n \ \Rightarrow\ n=10
Now computing {}^{10}C_{2}:
{}^{10}C_{2}=\dfrac{10!}{2!\,8!}=\dfrac{10\times9}{2}=45
Writing the ratio using the combination formula:
\dfrac{{}^{2n}C_{3}}{{}^{n}C_{3}}=\dfrac{\dfrac{2n!}{3!(2n-3)!}}{\dfrac{n!}{3!(n-3)!}}=\dfrac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}
Simplifying (cancelling n(n-1) and noting 2n-2=2(n-1)):
=\dfrac{2n(2n-1)\times2(n-1)}{n(n-1)(n-2)}=\dfrac{4(2n-1)}{n-2}
\dfrac{4(2n-1)}{n-2}=12 \ \Rightarrow\ 4(2n-1)=12(n-2)
8n-4=12n-24 \ \Rightarrow\ 20=4n \ \Rightarrow\ n=5
\dfrac{4(2n-1)}{n-2}=11 \ \Rightarrow\ 4(2n-1)=11(n-2)
8n-4=11n-22 \ \Rightarrow\ 18=3n \ \Rightarrow\ n=6
A chord is determined by choosing any 2 of the 21 points, and the order in which the two points are chosen does not matter — this is a selection, not an arrangement.
{}^{21}C_{2}=\dfrac{21!}{2!\,19!}=\dfrac{21\times20}{2}=210
3 boys can be selected from 5 boys in {}^{5}C_{3} ways, and independently, 3 girls can be selected from 4 girls in {}^{4}C_{3} ways.
{}^{5}C_{3}=\dfrac{5!}{3!\,2!}=10 \qquad {}^{4}C_{3}=\dfrac{4!}{3!\,1!}=4
By the multiplication principle, the required number of ways is:
10\times4=40
3 red balls can be chosen from 6 red balls, 3 white balls from 5 white balls, and 3 blue balls from 5 blue balls — each selection is independent of the others.
{}^{6}C_{3}=20 \qquad {}^{5}C_{3}=10 \qquad {}^{5}C_{3}=10
By the multiplication principle, the required number of ways is:
20\times10\times10=2000
A deck has 4 aces and 48 non-ace cards. Exactly one ace must be chosen from the 4 aces, and the remaining 4 cards of the hand must be chosen from the 48 non-ace cards.
{}^{4}C_{1}=4 \qquad {}^{48}C_{4}=\dfrac{48!}{4!\,44!}=194580
By the multiplication principle, the required number of combinations is:
4\times194580=778320
Exactly 4 bowlers must be chosen from the 5 players who can bowl, and the remaining 11-4=7 players of the team must be chosen from the 17-5=12 players who cannot bowl.
{}^{5}C_{4}=5 \qquad {}^{12}C_{7}=\dfrac{12!}{7!\,5!}=792
By the multiplication principle, the required number of ways is:
5\times792=3960
2 black balls can be chosen from 5 black balls, and independently, 3 red balls can be chosen from 6 red balls.
{}^{5}C_{2}=10 \qquad {}^{6}C_{3}=20
By the multiplication principle, the required number of ways is:
10\times20=200
Since 2 specific courses are compulsory, they are already part of every student's programme. Only the remaining 5-2=3 courses need to be chosen, and they must come from the 9-2=7 courses left over.
{}^{7}C_{3}=\dfrac{7!}{3!\,4!}=35
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