If nC8 = nC2, find nC2.
Hint. Use the fact that nCa=nCb implies either a=b or a+b=n.
Since 8 does not equal 2, we need n=8+2=10. Then nC2 = 10C2 = 45.
✦ Working through each part gives: 45.
Determine n if (i) 2nC3 : nC3 = 12:1, (ii) 2nC3 : nC3 = 11:1.
Hint. Write both combination terms in factorial form, simplify the ratio to a function of n alone, and solve the resulting linear equation.
The ratio 2nC3/nC3 simplifies to 4(2n-1)/(n-2). (i) Setting this to 12: 4(2n-1)=12(n-2), giving 8n-4=12n-24, so 4n=20, n=5. (ii) Setting it to 11: 4(2n-1)=11(n-2), giving 8n-4=11n-22, so 3n=18, n=6.
✦ Working through each part gives: (i) n = 5. (ii) n = 6.
How many chords can be drawn through 21 points on a circle?
Hint. Each chord corresponds to an unordered pair of points.
A chord is fully determined by an unordered pair of the 21 points, with order not mattering: 21C2 = (21x20)/2 = 210.
✦ Working through each part gives: 210.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Hint. Select the boys and girls independently, then multiply, since both selections happen together.
Selecting 3 boys from 5: 5C3=10. Selecting 3 girls from 4: 4C3=4. Since both selections happen together, multiply: 10x4=40.
✦ Working through each part gives: 40.
Find the number of ways of selecting 9 balls from 6 red, 5 white, and 5 blue balls if each selection consists of 3 balls of each colour.
Hint. Select 3 of each colour independently, then multiply.
6C3 x 5C3 x 5C3 = 20 x 10 x 10 = 2000.
✦ Working through each part gives: 2000.
Determine the number of 5-card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Hint. Select the one ace and the remaining four non-ace cards independently, then multiply.
1 ace from 4: 4C1=4. Remaining 4 cards from the 48 non-ace cards: 48C4=194580. Total: 4x194580=778320.
✦ Working through each part gives: 778320.
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Hint. Select the bowlers and the remaining non-bowling players independently, then multiply.
4 bowlers from 5: 5C4=5. Remaining 7 players from the 12 non-bowlers: 12C7=792. Total: 5x792=3960.
✦ Working through each part gives: 3960.
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Hint. Select black balls and red balls independently, then multiply.
Selecting 2 black from 5: 5C2=10. Selecting 3 red from 6: 6C3=20. Multiplying, since both selections happen together: 10x20=200.
✦ Working through each part gives: 200.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Hint. The 2 compulsory courses are already decided; only the remaining course slots need to be chosen from what's left.
With 2 courses fixed, 3 more must be chosen from the remaining 7 courses: 7C3=35.
✦ Working through each part gives: 35.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh106.pdf) — four numbered exercises (6.1-6.4, 31 questions) plus the chapter's Miscellaneous Exercise (11 questions); confirmed circular permutations do not appear anywhere in the chapter's theorems or its 42 questions, contrary to an earlier stub's claim about Exercise 6.3. Exercise 6.3 Q7's two-part nPr equation was solved with explicit domain checks (r must not exceed the smaller base number in each nPr term), rejecting an algebraically valid but combinatorially meaningless extraneous root in each part. Questions are referenced from the NCERT textbook for identification.
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