NCERT Solutions: 6.4 - Permutations And Combinations

Q1: If 

, find .
Ans: It is known that,
Therefore, substituting this known relation into the required expression gives
 
Hence, after simplification we obtain
 
This shows the required value obtained directly by replacingintoand simplifying.

Therefore,

 

 

Q2: Determine n if  (i)  

    (ii)

Ans: (i)

 

For part (i), equate the two expressions and simplify the equation step by step to solve for n.
(ii)

 

Equate the given expressions and rearrange terms to collect like terms:
⇒ 11n - 8n = - 4 + 22
⇒ 3n = 18
⇒ n = 6
Thus, n = 6 for part (ii).

Q3: How many chords can be drawn through 21 points on a circle?
Ans: For drawing 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 =

C(21, 2) = (21 × 20) ÷ 2 = 21 × 10 = 210.
So, 210 chords can be drawn.

Q4: In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Ans: 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 ways.
3 girls can be selected from 4 girls in ways.
Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected
 
Compute the values: C(5, 3) = (5 × 4 × 3) ÷ (3 × 2 × 1) = 10 and C(4, 3) = 4.
Number of ways = 10 × 4 = 40.

Q5: 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.
Ans: 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.
Here,
3 balls can be selected from 6 red balls in ways.
3 balls can be selected from 5 white balls in ways.
3 balls can be selected from 5 blue balls in ways.
Thus, by multiplication principle, required number of ways of selecting 9 balls

 

Now compute the numerical values: C(6, 3) = (6 × 5 × 4) ÷ (3 × 2 × 1) = 20 and C(5, 3) = 10.
Required number = 20 × 10 × 10 = 2000.

Q6: Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Ans: 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 ways and the remaining 4 cards can be selected out of the 48 cards in ways.
Thus, by multiplication principle, required number of 5 card combinations

Compute the values: C(4, 1) = 4 and C(48, 4) = (48 × 47 × 46 × 45) ÷ (4 × 3 × 2 × 1) = 194,580.
Required number = 4 × 194,580 = 778,320.

Q7: 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?
Ans: 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 ways and the remaining 7 players can be selected out of the 12 players in ways.
Thus, by multiplication principle, required number of ways of selecting cricket team
Compute the values: C(5, 4) = 5 and C(12, 7) = C(12, 5) = 792.
Required number = 5 × 792 = 3,960.

Q8: 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.
Ans: There are 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in

 ways and 3 red balls can be selected out of 6 red balls in ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls
Compute the values: C(5, 2) = 10 and C(6, 3) = 20.
Required number = 10 × 20 = 200.

Q9: 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?
Ans: 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 ways.
Thus, required number of ways of choosing the programme
Compute the value: C(7, 3) = (7 × 6 × 5) ÷ (3 × 2 × 1) = 35.
Therefore, the student can choose the programme in 35 ways.