Class 11 RD Sharma Solutions – Chapter 17 Combinations- Exercise 17.1 | Set 2

Question 11. If ²⁸C_2r : ²⁴C_2r-4=225:11, find r.

Solution:

(28!/(28-2r)!2r!)/(24!/(24-2r+4)!(2r-4)!)=225/11

28x27x26x25/2r(2r-1)(2r-2)(2r-3)=225/11

28x27x26x25x11=225x(2r)(2r-1)(2r-2)(2r-3)

28x3x26x11=2r(2r-1)(2r-2)(2r-3)

14x2x3x13x2x11=2r(2r-1)(2r-2)(2r-3)

14x13x12x11=2r(2r-1)(2r-2)(2r-3)

2r=14

r=7

Question 12. If ⁿC₄,ⁿC₅ and ⁿC₆ are in AP, then find n.

Solution:

We know that the A.P series is represented as a,a+d,a+2d,..

=>2.ⁿC₅=ⁿC₄+ⁿC₆

=>2(n!/(n-5)!5!)=(n!/(n-4)!4!)+n!/(n-6)!6!

=>2/(n-5)5=(1/(n-5)(n-4))+(1/30)

=>(2/(n-5)5)-(1/(n-5)(n-4))=1/30

=>(2(n-4)-5)30=(n-5)(n-4)5

=>60n-240-150=(n-5)(n-4)5

=>12n-78=(n-5)(n-4)

=>12n-78=n²-9n+20

=>n²-21n+98=0

=>n²-7n-14n+98=0

=>n(n-7)-14(n-7)=0

=>(n-7)(n-14)=0

=>n=7, 14

Question 13. If ²ⁿC₃:ⁿC₂=44:3, find n.

Solution:

(2n!/(2n-3)!3!)/(n!/(n-2)!2!)=44/3

2n(2n-1)(2n-2)/3n(n-1)=44/3

2n(2n-1)(2n-2)=44n(n-1)

(2n-1)(2n-2)=22n-22

4n²-6n+2=22n-22

4n²-28n+24=0

n²-7n+6=0

(n-1)(n-6)=0

n=1 , 6

Let n=1

then ²⁽¹⁾C₃:²C₂ is not possible because n<r.

So, n=6.

Question 14. If ¹⁶C_r=¹⁶C_r+2, find ^rC₄.

Solution:

16=r+r+2

16=2r+2

14=2r

r=7

=>⁷C₄=7!/3!4!

=5x6x7/3×2

=35

Question 15. If ∝=^mC₂, then find the value of ^∝C₂.

Solution:

^∝C₂=∝!/(∝-2)!2!

=(∝-1)(∝)/2

=(^mC₂-1)(^mC₂)/2

=(m!/(m-1)!2!-1)(m!/(m-2)!2!)/2

=(m(m-1)-2)(m(m-1))/8

=(m²-m-2)(m(m-1))/8

=(m+1)(m-1)(m-2)m/8

=1/8[(m-2)(m-1)(m)(m+1)]

Question 16. Prove that the product of 2n consecutive negative integers is divisible by (2n)!.

Solution:

Let the 2n consecutive negative integers be -k,-k-1,-k-2,…,-k-(2n-1)

product of 2n consecutive negative integers=-kx-k-1x-k-2x….x-k-(2n-1)

=(-1)²ⁿxkxk+1xk+2x…..xk+(2n-1)

=(-1)²ⁿxkxk+1xk+2x…..xk+(2n-1)x(k-1)!/(k-1)!

=(-1)²ⁿx(k+2n-1)!/(k-1)!

=(-1)²ⁿx(k+2n-1)!(2n)!/(k-1)!(2n)!

=(-1)²ⁿx^k+2n-1C_2nx(2n)!

Therefore the product of 2n consecutive negative integers is divided by (2n)!.

Question 17. For all positive integers n, show that ²ⁿC_n+²ⁿC_n-1=1/2[²ⁿ⁺²C_n+1].

Solution:

LHS=²ⁿC_n+²ⁿC_n-1

=2n!/n!(2n-n)!+2n!/(n-1)!(2n-n+1)!

=2n!/n!n!+2n!/(n-1)!(n+1)!

=2n!/n(n-1)!n!+2n!/(n-1)!n!(n+1)

=2n!/n(n+1)(n-1)!n![n+1+n]

=2n!(2n+1)/n(n+1)(n-1)!n!

=(2n+1)!/n!(n+1)!

=(2n+2)(2n+1)!/n!(n+1)!(2n+2)

=(2n+2)!/n!(n+1)!(n+1)2

=(2n+2)!/(n+1)!(n+1)!2

=(2n+2)!/(n+1)!(2n+2-n-1)!2

=1/2[²ⁿ⁺²C_n+1]

= RHS

Question 18. Prove that: ⁴ⁿC_2n:²ⁿC_n=[1x 3 x 5 x ….. x 4n-1]:[1 x 3 x 5 x …… x 2n-1]².

Solution:

LHS=⁴ⁿC_2n/²ⁿC_n

=(4n!/2n!2n!)/(2n!/n!n!)

=[1 x 2 x ……x 4n] x[1 x2x3x….xn]²/[1x2x…x2n]³

=[1x3x5x…4n-1][2x4x6x…4n](n!)(n!)/[1x3x5x…x2n-1]²[2x4x6x…x2n]²(2n!)

=[1x3x5x..x4n-1]2²ⁿ[1x2x3x..2n](n!)(n!)/[1x3x5x…x2n-1]²x2²ⁿ[1x2x3x..xn]²(2n)!

=[1x3x5x…x4n-1]/[1x3x5x2n-1]²

=RHS

Question 19. Evaluate

Solution:

=>²⁰C₅+²⁰C₄+²¹C₄+²²C₄+²³C₄

W.K.T ⁿC_r+ⁿC_r-1=ⁿ⁺¹C_r

=>²¹C₅+²¹C₄+²²C₄+²³C₄

=>²²C₅+²²C₄+²³C₄

=>²³C₅+²³C₄

=> ²⁴C₅

Question 20.1. Let r and n be positive integers such that 1<=r<=n. Then prove the following:

ⁿC_r/ⁿC_r-1=n-r+1/r

Solution:

LHS=(n!/r!(n-r)!)/(n!/(r-1)!(n-r+1)!

=(n-r)!(n-r+1)(r-1)!/(n-r)!(r)(r-1)!

=n-r+1/r

=RHS

Question 20.2. Let r an dn be positive integers such that 1<=r<=n. Then prove the following:

nx^n-1C_r-1=(n-r+1)ⁿC_r-1

Solution:

LHS=n(n-1)!/(r-1)!(n-1-r+1)!

=n!/(r-1)!( n-r)!

=n!(n-r+1)/(r-1)!( n-r+1)(n-r)!

=n!(n-r+1)/(r-1)!(n-r+1)!

=(n-r+1)ⁿC_r-1

=RHS

Question 20.3. Let r and n be positive integers such that 1<=r<=n. Then prove the following:

ⁿC_r/^n-1C_r-1=n/r

Solution:

LHS=(n!/r!(n-r)!)/(n-1)!/(n-r)!(r-1)!

=n(n-1)!(r-1)!(n-r)!/r(r-1)!(n-r)!(n-1)!

=n/r

=RHS

Question 20.4. Let r and n be positive integers such that 1<=r<=n. Then prove the following:

ⁿC_r+2 x ⁿC_r-1+ ⁿC_r-2=ⁿ⁺²C_r.

Solution:

W.K.T ⁿC_r+ⁿC_r-1=ⁿ⁺¹C_r

LHS=ⁿC_r+ⁿC_r-1+ⁿC_r-1+ⁿC_r-2

=ⁿ⁺¹C_r+ⁿ⁺¹C_r-1

=ⁿ⁺²C_r

=RHS

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