1. If $^{n}C_{8}=^{n}C_{2}$, find $^{n}C_{2}$
We know that,
if $^{n}C_{a}=^{n}C_{b}$, a=b or n=a+b
$^{n}C_{8}=^{n}C_{2}$
⇒ $n=a+b$
⇒ $n=8+2=10$
$^{n}C_{2}=^{10}C_{2}$
= $\frac{10!}{2!8!}$
= $\frac{10\times 9\times 8!}{2!8!}$
= $\frac{10\times 9}{2}$
= $45$
2. Determine n if
(i) $^{2n}C_{3}:^{n}C_{3}$ $=12:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=12:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=12:1$
$\large \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}$ $=\frac{12}{1}$
$\large \frac{\frac{(2n)(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!}}{\frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}}$ $=12$
$\large \frac{\frac{2n(2n-1)(2n-2)}{3!}}{\frac{n(n-1)(n-2)}{3!}}$ $=12$
$\large \frac{2n(2n-1)(2n-2)}{3!}*\frac{3!}{n(n-1)(n-2)}$ $=12$
$\large \frac{2(2n-1)(2n-2)}{(n-1)(n-2)}$ $=12$
$\large \frac{2(2n-1)2(n-1)}{(n-1)(n-2)}$ $=12$
$\large \frac{4(2n-1)}{(n-2)}$ $=12$
$4(2n-1)=12(n-2)$
$8n-4=12n-24$
$-4+24=12n-8n$
$20=4n$
$n=5$
(ii) $^{2n}C_{3}:^{n}C_{3}$ $=11:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=11:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=11:1$
$\large \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}$ $=\frac{11}{1}$
$\large \frac{\frac{(2n)(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!}}{\frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}}$ $=11$
$\large \frac{\frac{2n(2n-1)(2n-2)}{3!}}{\frac{n(n-1)(n-2)}{3!}}$ $=11$
$\large \frac{2n(2n-1)(2n-2)}{3!}*\frac{3!}{n(n-1)(n-2)}$ $=11$
$\large \frac{2(2n-1)(2n-2)}{(n-1)(n-2)}$ $=11$
$\large \frac{2(2n-1)2(n-1)}{(n-1)(n-2)}$ $=11$
$\large \frac{4(2n-1)}{(n-2)}$ $=11$
$4(2n-1)=11(n-2)$
$8n-4=11n-22$
$-4+22=11n-8n$
$18=3n$
$n=6$
3. How many chords can be drawn through 21 points on a circle?
For drawing one chord on a circle, 2 points are required.
To know the number of chords that can be drawn through 21 points, the number of combinations have to be counted.
Therefore, combinations of 21 points taken 2 at a time is $^{21}C_{2}$
$^{21}C_{2}=\frac{21!}{2!(21-2)!}$
= $\frac{21!}{2!19!}$
= $\frac{21\times 20\times 19!}{2!19!}$
= $\frac{21\times 20}{2!}$
= $210$
4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
A team of 3 boys and 3 girls to be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in $^{5}C_{3}$ ways.
3 girls can be selected from 4 girls in $^{4}C_{3}$ ways.
Therefore, by multiplication principle, number of combinations is $^{5}C_{3}*^{4}C_{3}$
$^{5}C_{3}*^{4}C_{3}$ $=\frac{5!}{3!(5-3)!}*\frac{4!}{3!(4-3)!}$
= $\frac{5!}{3!2!}*\frac{4!}{3!1!}$
= $\frac{5\times 4\times 3!}{3!2!}*\frac{4\times 3!}{3!}$
= $\frac{5\times 4}{2}*4$
= $40$
5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
There are a total of 6 red balls, 5 white balls and 5 blue balls.
9 balls to be selected in such a way that each selection consists of 3 balls of each colour.
3 balls can be selected from 6 red balls in $^{6}C_{3}$ ways.
3 balls can be selected from 5 white balls in $^{5}C_{3}$ ways.
3 balls can be selected from 5 blue balls in $^{5}C_{3}$ ways.
Therefore, by multiplication principle, number of combinations is $^{6}C_{3}*^{5}C_{3}*^{5}C_{3}$
= $\frac{6!}{3!(6-3)!}*\frac{5!}{3!(5-3)!}*\frac{5!}{3!(5-3)!}$
= $\frac{6\times 5\times 4\times 3!}{3!3!}$ $*\frac{5\times 4\times 3!}{3!2!}$ $*\frac{5\times 4\times 3!}{3!2!}$
= $\frac{6\times 5\times 4}{3!}$ $*\frac{5\times 4}{2!}$ $*\frac{5\times 4}{2!}$
= $\frac{6\times 5\times 4}{6}$ $*\frac{5\times 4}{2}$ $*\frac{5\times 4}{2}$
= $5\times 4\times 5\times 2\times 5\times 2$
= $2000$
6. Find the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
In a deck of 52 cards, there are 4 aces.
A combination of 5 cards have to be made in which there is exactly one ace.
Then, 1 ace can be selected out of 4 ace in $^{4}C_{1}$ ways.
remaining 4 cards can be selected out of 48 cards in $^{48}C_{4}$ ways.
Therefore, by multiplication principle, number of combinations is $^{4}C_{1}*^{48}C_{4}$
$^{4}C_{1}*^{48}C_{4}$ $=\frac{4!}{1!(4-1)!}*\frac{48!}{4!(48-4)!}$
= $\frac{48!}{3!44!}$
= $\frac{48\times 47\times 46\times 45\times 44!}{3!44!}$
= $\frac{48\times 47\times 46\times 45}{6}$
= $8\times 47\times 46\times 45$
= $778320$
7. 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?
11 players to be selected such that there are exactly 4 bowlers.
4 bowlers can be selected out of 5 bowlers in $^{5}C_{4}$ ways.
remaining 7 players can be selected out of players in $^{12}C_{7}$ ways.
Therefore, by multiplication principle, number of combinations is $^{5}C_{4}*^{12}C_{7}$
= $\frac{5!}{4!(5-4)!}*\frac{12!}{7!(12-7)!}$
= $\frac{5!}{4!1!}*\frac{12!}{7!5!}$
= $\frac{12\times11\times10\times 9\times 8\times 7!}{4!7!}$
= $\frac{12\times11\times10\times 9\times 8}{24}$
= $3960$
8. 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.
There are a total of 5 black balls and 6 red balls.
2 balls can be selected from 5 black balls in $^{5}C_{2}$ ways.
3 balls can be selected from 6 red balls in $^{6}C_{3}$ ways.
Therefore, by multiplication principle, number of combinations is $^{5}C_{2}*^{6}C_{3}$
= $\frac{5!}{2!(5-2)!}*\frac{6!}{3!(6-3)!}$
= $\frac{5\times 4\times 3!}{2!3!}*\frac{6\times 5\times 4\times 3!}{3!3!}$
= $\frac{5\times 4\times 3!}{2!3!}*\frac{6\times 5\times 4\times 3!}{3!3!}$
= $\frac{5\times 4}{2!}*\frac{6\times 5\times 4}{3!}$
= $\frac{5\times 4}{2}*\frac{6\times 5\times 4}{6}$
= $200$
9. In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
5 courses to be selected from 9 courses.
2 courses are compulsory.
Remaining 3 courses to be selected from remaining 7 courses $^{7}C_{3}$
= $\frac{7!}{3!(7-3)!}$
= $\frac{7!}{3!4!}$
= $\frac{7\times 6\times 5\times 4!}{3!4!}$
= $\frac{7\times 6\times 5}{6}$
= $35$
We know that,
if $^{n}C_{a}=^{n}C_{b}$, a=b or n=a+b
$^{n}C_{8}=^{n}C_{2}$
⇒ $n=a+b$
⇒ $n=8+2=10$
$^{n}C_{2}=^{10}C_{2}$
= $\frac{10!}{2!8!}$
= $\frac{10\times 9\times 8!}{2!8!}$
= $\frac{10\times 9}{2}$
= $45$
2. Determine n if
(i) $^{2n}C_{3}:^{n}C_{3}$ $=12:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=12:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=12:1$
$\large \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}$ $=\frac{12}{1}$
$\large \frac{\frac{(2n)(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!}}{\frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}}$ $=12$
$\large \frac{\frac{2n(2n-1)(2n-2)}{3!}}{\frac{n(n-1)(n-2)}{3!}}$ $=12$
$\large \frac{2n(2n-1)(2n-2)}{3!}*\frac{3!}{n(n-1)(n-2)}$ $=12$
$\large \frac{2(2n-1)(2n-2)}{(n-1)(n-2)}$ $=12$
$\large \frac{2(2n-1)2(n-1)}{(n-1)(n-2)}$ $=12$
$\large \frac{4(2n-1)}{(n-2)}$ $=12$
$4(2n-1)=12(n-2)$
$8n-4=12n-24$
$-4+24=12n-8n$
$20=4n$
$n=5$
(ii) $^{2n}C_{3}:^{n}C_{3}$ $=11:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=11:1$
$\frac{(2n)!}{3!(2n-3)!}:\frac{n!}{3!(n-3)!}$ $=11:1$
$\large \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}$ $=\frac{11}{1}$
$\large \frac{\frac{(2n)(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!}}{\frac{n(n-1)(n-2)(n-3)!}{3!(n-3)!}}$ $=11$
$\large \frac{\frac{2n(2n-1)(2n-2)}{3!}}{\frac{n(n-1)(n-2)}{3!}}$ $=11$
$\large \frac{2n(2n-1)(2n-2)}{3!}*\frac{3!}{n(n-1)(n-2)}$ $=11$
$\large \frac{2(2n-1)(2n-2)}{(n-1)(n-2)}$ $=11$
$\large \frac{2(2n-1)2(n-1)}{(n-1)(n-2)}$ $=11$
$\large \frac{4(2n-1)}{(n-2)}$ $=11$
$4(2n-1)=11(n-2)$
$8n-4=11n-22$
$-4+22=11n-8n$
$18=3n$
$n=6$
3. How many chords can be drawn through 21 points on a circle?
For drawing one chord on a circle, 2 points are required.
To know the number of chords that can be drawn through 21 points, the number of combinations have to be counted.
Therefore, combinations of 21 points taken 2 at a time is $^{21}C_{2}$
$^{21}C_{2}=\frac{21!}{2!(21-2)!}$
= $\frac{21!}{2!19!}$
= $\frac{21\times 20\times 19!}{2!19!}$
= $\frac{21\times 20}{2!}$
= $210$
4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
A team of 3 boys and 3 girls to be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in $^{5}C_{3}$ ways.
3 girls can be selected from 4 girls in $^{4}C_{3}$ ways.
Therefore, by multiplication principle, number of combinations is $^{5}C_{3}*^{4}C_{3}$
$^{5}C_{3}*^{4}C_{3}$ $=\frac{5!}{3!(5-3)!}*\frac{4!}{3!(4-3)!}$
= $\frac{5!}{3!2!}*\frac{4!}{3!1!}$
= $\frac{5\times 4\times 3!}{3!2!}*\frac{4\times 3!}{3!}$
= $\frac{5\times 4}{2}*4$
= $40$
5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
There are a total of 6 red balls, 5 white balls and 5 blue balls.
9 balls to be selected in such a way that each selection consists of 3 balls of each colour.
3 balls can be selected from 6 red balls in $^{6}C_{3}$ ways.
3 balls can be selected from 5 white balls in $^{5}C_{3}$ ways.
3 balls can be selected from 5 blue balls in $^{5}C_{3}$ ways.
Therefore, by multiplication principle, number of combinations is $^{6}C_{3}*^{5}C_{3}*^{5}C_{3}$
= $\frac{6!}{3!(6-3)!}*\frac{5!}{3!(5-3)!}*\frac{5!}{3!(5-3)!}$
= $\frac{6\times 5\times 4\times 3!}{3!3!}$ $*\frac{5\times 4\times 3!}{3!2!}$ $*\frac{5\times 4\times 3!}{3!2!}$
= $\frac{6\times 5\times 4}{3!}$ $*\frac{5\times 4}{2!}$ $*\frac{5\times 4}{2!}$
= $\frac{6\times 5\times 4}{6}$ $*\frac{5\times 4}{2}$ $*\frac{5\times 4}{2}$
= $5\times 4\times 5\times 2\times 5\times 2$
= $2000$
6. Find the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
In a deck of 52 cards, there are 4 aces.
A combination of 5 cards have to be made in which there is exactly one ace.
Then, 1 ace can be selected out of 4 ace in $^{4}C_{1}$ ways.
remaining 4 cards can be selected out of 48 cards in $^{48}C_{4}$ ways.
Therefore, by multiplication principle, number of combinations is $^{4}C_{1}*^{48}C_{4}$
$^{4}C_{1}*^{48}C_{4}$ $=\frac{4!}{1!(4-1)!}*\frac{48!}{4!(48-4)!}$
= $\frac{48!}{3!44!}$
= $\frac{48\times 47\times 46\times 45\times 44!}{3!44!}$
= $\frac{48\times 47\times 46\times 45}{6}$
= $8\times 47\times 46\times 45$
= $778320$
7. 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?
11 players to be selected such that there are exactly 4 bowlers.
4 bowlers can be selected out of 5 bowlers in $^{5}C_{4}$ ways.
remaining 7 players can be selected out of players in $^{12}C_{7}$ ways.
Therefore, by multiplication principle, number of combinations is $^{5}C_{4}*^{12}C_{7}$
= $\frac{5!}{4!(5-4)!}*\frac{12!}{7!(12-7)!}$
= $\frac{5!}{4!1!}*\frac{12!}{7!5!}$
= $\frac{12\times11\times10\times 9\times 8\times 7!}{4!7!}$
= $\frac{12\times11\times10\times 9\times 8}{24}$
= $3960$
8. 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.
There are a total of 5 black balls and 6 red balls.
2 balls can be selected from 5 black balls in $^{5}C_{2}$ ways.
3 balls can be selected from 6 red balls in $^{6}C_{3}$ ways.
Therefore, by multiplication principle, number of combinations is $^{5}C_{2}*^{6}C_{3}$
= $\frac{5!}{2!(5-2)!}*\frac{6!}{3!(6-3)!}$
= $\frac{5\times 4\times 3!}{2!3!}*\frac{6\times 5\times 4\times 3!}{3!3!}$
= $\frac{5\times 4\times 3!}{2!3!}*\frac{6\times 5\times 4\times 3!}{3!3!}$
= $\frac{5\times 4}{2!}*\frac{6\times 5\times 4}{3!}$
= $\frac{5\times 4}{2}*\frac{6\times 5\times 4}{6}$
= $200$
9. In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
5 courses to be selected from 9 courses.
2 courses are compulsory.
Remaining 3 courses to be selected from remaining 7 courses $^{7}C_{3}$
= $\frac{7!}{3!(7-3)!}$
= $\frac{7!}{3!4!}$
= $\frac{7\times 6\times 5\times 4!}{3!4!}$
= $\frac{7\times 6\times 5}{6}$
= $35$
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