- If nC8 = nC2. find nC2
Solution:
We know that, nCa = nCb then a = b or n = a + b
Therefore,
nC8 = nC2, since 8≠2 , we have 8 + 2 = 10
nC2 = 10C2 = 10!/2!(10-2)! = 10!/2!8! = 10x9x8/2x1x8 = 45
- Determine n if
(i) (2nC3 : nC3 ) = 12 : 1
(ii) 2nC3 : nC3 = 11 : 1
Solution:
(i) (2nC3 : nC3 ) = 12 : 1
(ii) 2nC3 : nC3 = 11 : 1
- How many chords can be drawn through 21 points on a circle?
solution:
We know that, to draw one chord on a circle, only 2 points are required.
To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.
Therefore, there will be as many chords as there are combinations of 21 points taken 2 at a time.
Thus, required number of chords = 21C2 = 21!/2!(21-2)! = 21!/21!9! = 21×20/2 = 210
- In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Solution:
It is given that, a team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in 5C3 ways.
3 girls can be selected from 4 girls in 4C3 ways.
Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected = 5C3 x 4C3 = 5!/3!2! x 4!/3!1!
= 5x4x3!/3! x 4×3!/3!
= 10 x 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.
Solution:
It is given that, there are a total of 6 red balls, 5 white balls, and 5 blue balls. 9 balls have 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 6C3 ways.
3 balls can be selected from 5 white balls in 5C3 ways.
3 balls can be selected from 5 blue balls in 5C3 ways.
Thus, by multiplication principle, required number of ways of selecting 9 balls
= 6C3 x 5C3 x 5C3
= 6!/3!3! x 5!/3!2! x 5!/3!2!
= 6x5x4x3x2x1/3!x3x2 x 5x4x3x2x1/3!x2 x 5x4x3x2x1/3!x2
= 20 x 10 x 10
= 2000
- Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
solution:
IT is given that, 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, one ace can be selected in 4C1 ways and the remaining 4 cards can be selected out of the 48 cards in 48C4 ways.
Thus, by multiplication principle, required number of 5 card combinations
48C4 x 4C1 = 48!/4!44! x 4!/1!3!
= 48x47x46x45/4x3x2x1 x 4
= 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?
Solution:
It is given that out of 17 players, 5 players are bowlers.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
4 bowlers can be selected in 5C4 ways and the remaining 7 players can be selected out of the 12 players in 12C7 ways.
Thus, by multiplication principle, required number of ways of selecting cricket team = 5C4 x 12C7 = 5!/4!1! x 12!/7!5! = 5 x 12x11x10x9x8/5x4x3x2x1 = 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.
solution:
We have 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in 5C2 ways and 3 red balls can be selected out of 6 red balls in 6C3 ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls
5C2 x 6C3 = 5!/2!3! x 6!/3!3! = 5×4/2 x 6x5x4/3x2x1 = 10 x20 = 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?
solution:
There are 9 courses available out of which, 2 specific courses are compulsory for every student.
Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in 7C3 ways.
Thus the required number of ways of choosing the programme
= 7C3 = 7!/3!4! = 7x6x5x4!/3x2x1x4 = 35
