Class 11 Maths Ncert Solutions Chapter 7 Ex 7.4 Permutations & Combinations PDF

# Class 11 Maths Ncert Solutions Ex 7.4

## Class 11 Maths Ncert Solutions Chapter 7 Ex 7.4

Q-1: If nC8=nC3$^{ n }C_{ 8 } = ^{ n }C_{ 3 }$, find the value of nC3$^{ n }C_{ 3 }$

Solution:

We know that, if nCx=nCy$^{ n }C_{ x } = \;^{ n }C_{ y }$

So, x = y or n = x + y

Hence, nC8=nC3$^{ n }C_{ 8 } = \;^{ n }C_{ 3 }$

n = 8 + 3 = 11

Therefore, nC2=11C2$^{ n }C_{ 2 } = \;^{ 11 }C_{ 2 }$=11!2!(112)!$\frac{ 11! }{ 2!\left ( 11 \;-\; 2 \right )!}$ = 11×10×9!2×1×9!$\frac{ 11\; \times \;10\; \times \;9! }{ 2\; \times \;1 \;\times \;9! }$ = 11×102$\frac{ 11\; \times \;10 }{ 2 }$

Therefore, nC8=nC3$^{ n }C_{ 8 } = \;^{ n }C_{ 3 }$ = 55

Q-2: Find the value of ‘a’ if:

(i) 2aC2:aC2$^{ 2a }C_{ 2 } : \;^{ a }C_{ 2 }$ = 12 : 1

(ii) 2aC2:aC2$^{ 2a }C_{ 2 } : \;^{ a }C_{ 2 }$ = 11 : 1

Solution:

(i). 2aC2aC2=121$\frac{ ^{2a }C_{ 2 } }{ ^{ a }C_{ 2 } } = \frac{ 12 }{ 1 }\\$

$\\\Rightarrow$  (2a)!2!(2a2)!×2!(2a2)!a!=121$\frac{\left ( 2a \right )! }{ 2!\left ( 2a\; -\; 2 \right )! } \times \frac{ 2!\left ( 2a\; -\; 2 \right )! }{ a! } = \frac{ 12 }{ 1 }\\$

$\\\Rightarrow$   (2a)(2a1)(2a2)!(2a2)!×(a2)!a(a1)(a2)!=12$\frac{ \left ( 2a \right )\left ( 2a \;- \;1 \right )\left ( 2a \;- \;2 \right )! }{ \left ( 2a\; -\; 2 \right )!} \times \frac{ \left ( a \;-\; 2 \right )! }{ a\left ( a\; – \;1 \right )\left ( a\; – \;2 \right )! } = 12\\$

$\\\Rightarrow$   (2)(2a1)(a1)(a2)=12$\frac{ \left ( 2 \right )\left ( 2a \;-\; 1 \right ) }{ \left ( a \;-\; 1 \right )\left ( a\; – \;2 \right ) } = 12\\$

$\\\Rightarrow$   4(a1)(2a1)(a1)(a2)=12$\frac{ 4 \left ( a \;-\; 1 \right )\left ( 2a \;-\; 1 \right ) }{ \left ( a \;-\; 1 \right )\left ( a \;-\; 2 \right ) } = 12\\$

$\\\Rightarrow$   (2a1)(a2)=3$\frac{ \left ( 2a \;- \;1 \right ) }{ \left ( a \;-\; 2 \right ) } = 3$

So, 2a – 1 = 3( a – 2 )

2a – 1 = 3a – 6

3a – 2a = -1 + 6

Therefore, the value of a = 5

(ii). 2aC2aC2=111$\frac{ ^{2a }C_{ 2 } }{ ^{ a }C_{ 2 } } = \frac{ 11 }{ 1 }\\$

$\\\Rightarrow$   (2a)!2!(2a2)!×2!(2a2)!a!=111$\frac{\left ( 2a \right )! }{ 2!\left ( 2a\; -\; 2 \right )! } \times \frac{ 2!\left ( 2a \;-\; 2 \right )! }{ a! } = \frac{ 11 }{ 1 }\\$

$\\\Rightarrow$   (2a)(2a1)(2a2)!(2a2)!×(a2)!a(a1)(a2)!=11$\frac{ \left ( 2a \right )\left ( 2a \;-\; 1 \right )\left ( 2a \;-\; 2 \right )! }{ \left ( 2a\; -\; 2 \right )!} \times \frac{ \left ( a\; -\; 2 \right )! }{ a\left ( a\; -\; 1 \right )\left ( a \;-\; 2 \right )! } = 11\\$

$\\\Rightarrow$   (2)(2a1)(a1)(a2)=11$\frac{ \left ( 2 \right )\left ( 2a \;- \;1 \right ) }{ \left ( a\; -\; 1 \right )\left ( a \;-\; 2 \right ) } = 11\\$

$\\\Rightarrow$   4(a1)(2a1)(a1)(a2)=11$\frac{ 4 \left ( a\; -\; 1 \right )\left ( 2a\; – \;1 \right ) }{ \left ( a\; – \;1 \right )\left ( a \;-\; 2 \right ) } = 11\\$

$\\\Rightarrow$   4(2a1)=11(a2)$4\left ( 2a \;- \;1 \right ) = 11\left ( a \;-\; 2 \right )$

So, 8a – 4 = 11( a – 2 )

8a – 4 = 11a – 22

11a – 8a = -4 + 22

3a = 18

Therefore, the value of a = 6

Q-3: Find total number of chords which can be drawn through 21 points on the circle.

Solution:

2 points are required to draw one chord on a circle.

Now, in order to obtain the number of chords which can be drawn through the given 21 points on the circle, we need to count total number of the combinations.

Hence, there can be several numbers of chords as there are combinations of 21 points which can be taken 2 at a time.

Therefore, number of chords required = 21C2=21!2!(212)!=21!2!19!=21×202$^{ 21 }C_{ 2 } = \frac{ 21! }{ 2!\left ( 21\; -\; 2 \right )! } = \frac{ 21! }{ 2! \;19! } = \frac{ 21 \;\times \;20 }{ 2 }$ = 210

Q-4: Find the number of ways in which a team of 2 boys and 2 girls can be selected from the group of 5 boys and 4 girls.

Solution:

We have to select 2 boys and 2 girls from a group of 5 boys and 4 girls to form a team.

Now,

2 boys should be selected among 5 boys group in 5C2$^{ 5 }C_{ 2 }$ ways.

Also, 2 girls should be selected among a group of 4 girls in 4C2$^{ 4 }C_{ 2 }$ ways.

Hence, by the principle of multiplication, the number of ways by which a team of 2 boys and 2 girls can be formed:

= 5C2×4C2$^{ 5 }C_{ 2 } \times ^{ 4 }C_{ 2 }$ = 5!2!3!×4!2!2!$\frac{ 5! }{ 2! \;3! } \times \frac{ 4! }{ 2! \;2! }$= 5×4×3!2×1×3!×4×3×2!2×1×2!$\frac{ 5\; \times 4\; \times \;3! }{ 2\; \times \;1\; \times \;3! } \times \frac{ 4\; \times \;3\; \times \;2! }{ 2\; \times \;1\; \times \;2! }$= 5 × 2 × 2 × 3 = 60 ways

Q-5: In how many ways will 9 balls be selected from 7 red balls, 6 blue balls and 5 white balls if every time 3 balls of different colors be selected?

Solution:

We have total of 6 blue balls, 5 white balls and 7 red balls.

We have to select 9 balls from these 18 balls in such a ways that every time 3 different colored balls can be selected.

Here,

We can select 3 balls from 7 red balls in 7C3$^{ 7 }C_{ 3 }$ number of ways.

We can select 3 balls from 6 blue balls in 6C3$^{ 6 }C_{ 3 }$ number of ways.

We can select 3 balls from 5 white balls in 5C3$^{ 5 }C_{ 3 }$ number of ways.

Therefore, by the principle of multiplication, number of ways required to select 9 balls:

= 7C3×6C3×5C3$^{ 7 }C_{ 3 } \times ^{ 6 }C_{ 3 } \times ^{ 5 }C_{ 3 }$ = 7!3!4!×6!3!3!×5!3!2!$\frac{ 7! }{ 3!\; 4! } \times \frac{ 6! }{ 3! \;3! } \times \frac{ 5! }{ 3!\; 2! }$ = 7×6×5×4!4!×3×2×1×6×5×4×3!3!×3×2×1×5×4×3!3!×2×1$\frac{ 7\; \times \;6\; \times \;5\; \times \;4! }{ 4!\; \times \;3\; \times \;2\; \times \;1 } \times \frac{ 6\; \times \;5\; \times \;4\; \times \;3! }{ 3!\; \times 3 \;\times 2\; \times \;1 } \times \frac{ 5\; \times \;4\; \times \;3! }{ \;3!\; \times \;2 \;\times \;1 }$

= 7000 ways

Q-6: Find out the number of combinations of 5 cards from a deck of 52 cards. Assume that there is at least one ace in each of the combinations.

Solution:

We know that, we have 52 cards in a deck where there are 4 aces.

We have to make a combination of 5 cards in which there will be one ace for sure.

As we have 4 aces in the deck so, aces can be selected in 4C1$^{ 4 }C_{ 1 }$ number of ways and out of 48 cards in the deck, 4 cards can be selected in 48C4$^{ 48 }C_{ 4 }$ number of ways.

Therefore, by the principle of the multiplication, number of the combination of 5 cards required:

= 48C4×4C1$^{ 48 }C_{ 4 } \times ^{ 4 }C_{ 1 }$= 48!4!44!×4!1!3!$\frac{ 48! }{ 4!\; 44! } \times \frac{ 4! }{ 1!\; 3! }$= 48×47×46×45×44!4×3×2×1×44!×4×3!3!$\frac{ 48\; \times 47\; \times \;46\; \times \;45\; \times \;44! }{ 4\; \times 3\; \times 2\; \times 1\; \times \;44! } \times \frac{ 4\; \times \;3! }{ 3! }$

= 48×47×46×454×3×2×1×4$\frac{ 48\; \times \;47\; \times \;46\; \times \;45 }{ 4\; \times \;3\; \times \;2\; \times \;1 } \times 4$

= 778320 number of combinations.

Q-7: Find the number of ways by which one can select a cricket team of 11 players from a bunch of 17 players among which only 5 players can bowl if every cricket team of 11 players must have exactly 4 bowlers.

Solution:

Among 17 players, 5 players can bowl.

Now, we need to select a cricket team of 11 players where there are exactly 4 players who can bowl.

So, Among 5, 4 bowlers can be selected in 5C4$^{ 5 }C_{ 4 }$ number of ways and the remaining 7 players for the team of 11 players will be selected from the remaining 12 players in 12C7$^{ 12 }C_{ 7 }$ number of ways.

Therefore, by the principle of multiplication, number of ways required to select a cricket team:

= 5C4×12C7$^{ 5 }C_{ 4 } \times ^{ 12 }C_{ 7 }$ = 5!4!1!×12!7!5!$\frac{ 5! }{ 4!\; 1! } \times \frac{ 12! }{ 7!\; 5! }\\$

= 5×4!4!×12×11×10×9×8×7!7!×5×4×3×2×1$\\\frac{ 5\; \times \;4! }{ 4! } \times \frac{ 12\; \times \;11\; \times \;10\; \times \;9\; \times \;8\; \times \;7! }{ 7!\; \times \;5\; \times \;4\; \times \;3\; \times \;2\; \times \;1 }\\$

= 5×12×11×10×9×85×4×3×2×1$\\\frac{ 5\; \times \;12\; \times \;11\; \times \;10\; \times \;9\; \times \;8 }{ 5\; \times \;4\; \times \;3\; \times \;2\; \times \;1 }\\$

= 3960 number of ways.

Q-8: There are 6 black and 5 red balls. Find the number of ways through which 3 black and 3 red balls will be selected.

Solution:

We have a bag which contains 6 black and 5 red balls.

3 black balls will be selected from 6 black balls in 6C3 number of ways and 3 red balls will be selected out of 5 red balls in 5C3 number of ways.

Therefore, by the principle of multiplication, required number of ways to select 3 black and 3 red balls:

= 6C3×5C3$^{ 6 }C_{ 3 } \times ^{ 5 }C_{ 3 }$= 6!3!3!×5!3!2!$\frac{ 6! }{ 3!\; 3! } \times \frac{ 5! }{ 3!\; 2! }$= 6×5×4×3!3!×3×2×1×5×4×3!3!×2×1$\frac{ 6\; \times \;5\; \times \;4\; \times \;3! }{ 3!\; \times \;3\; \times \;2\; \times \;1 } \times \frac{ 5\; \times \;4\; \times \;3! }{ 3!\; \times \;2\; \times \;1 }$= 5×4×5×2$5 \times 4 \times 5 \times 2$

= 200 number of ways.

Q-9: Find the number of ways by which a student can choose a program of 5 courses, if there are options of 10 courses and for every student 2 courses are made compulsory.

Solution:

10 courses are available out of which 2 courses are made compulsory for each student.

Hence, each student have to choose 3 courses from the remaining courses, i.e., 8 other remaining courses, which can be selected in 8C3$^{ 8 }C_{ 3 }$ number of ways.

Therefore, the total number of ways in which program can be chosen:

= 8C3$^{ 8 }C_{ 3 }$= 8!3!5!$\frac{ 8! }{ 3!\; 5! }$= 8×7×6×5!5!×3×2×1$\frac{ 8\; \times \;7\; \times \;6\; \times \;5! }{ 5!\; \times \;3\; \times \;2\; \times \;1 }$

= 56 number of ways.