HOW TO FIND THE VALUE OF N AND R IN COMBINATION

Question 1 :

If ⁿC₁₂ = ⁿC₉ find ²¹C_n.

Solution :

ⁿC₁₂  =  n!/(n - 12)! 12!  -----(1)

ⁿC₉  =  n!/(n - 9)! 9!     -----(2)

(1)  =  (2)  

n!/(n - 12)! 12!  =  n!/(n - 9)! 9!

(n - 9)! 9!  =  (n - 12)! 12!

(n - 9)(n - 10)(n - 11)(n - 12)! 9!  =  (n - 12)! 12 ⋅ 11 ⋅ 10 ⋅ 9!

(n - 9)(n - 10)(n - 11)  =  12 ⋅ 11 ⋅ 10 

n - 9  =  12

n  =  12 + 9

n  =  21

 ²¹C_n =   ²¹C₂₁  =  1

Hence the answer is 1.

Question 2 :

If ¹⁵C_2r−1 = ¹⁵C_2r+4, find r.

Solution :

If ⁿC_x  =  ⁿC_y  ==>  x  =  y (or) x + y  =  n

2r - 1 + 2r + 4  =  15

4r + 3  =  15

4r  =  12

Divide by 4 on both sides.

r  =  12/4  ==>  3

Hence the value of r is 3.

Question 3 :

If ⁿP_r = 720, and ⁿC_r = 120, find n, r.

Solution :

ⁿP_r = 720

n!/(n - r)!  =  720  ---(1)

ⁿC_r = 120

n!/(n - r)! r!  =  120  ---(2)

Divide (1) by (2), we get 

r!  =  720/120

r!  =  6

r!  =  3!  ==> r = 3

By applying the value of r in the (1), we get

n!/(n - 3)!  =  720 

n(n - 1) (n - 2)  =  720

n(n - 1) (n - 2)  =  10 ⋅ 9 ⋅ 8

n  =  10

Hence the value of r and n are 3 and 10 respectively.

Question 4 :

Prove that ¹⁵C₃ + 2 × ¹⁵C₄ + ¹⁵C₅ = ¹⁷C₅.

Solution :

L.H.S 

  =   ¹⁵C₃ + 2 × ¹⁵C₄ + ¹⁵C₅

  =   ¹⁵C₃ + ¹⁵C₄ + ¹⁵C₄ + ¹⁵C₅

=   ¹⁵C₄ + ¹⁵C₃ +  ¹⁵C₅ + ¹⁵C₄

By using the property :

ⁿC_r + ⁿ C_r−1 = ⁿ⁺¹C_r

=  ¹⁶C₄ + ¹⁶C₅

=  ¹⁷C₅ ---> R.H.S

Question 5 :

Prove that ³⁵C₅ + ∑ ⁴_r=0 ^(39−r)C₄  =  ⁴⁰C₅.

Solution :

L.H.S

 =  ³⁵C₅ + ∑ ⁴_r=0 ^(39−r)C₄

  =  ³⁵C₅ + ³⁹C₄+ ³⁸C₄ + ³⁷C₄ + ³⁶C₄ + ³⁵C₄

By using the property :

ⁿC_r + ⁿ C_r−1 = ⁿ⁺¹C_r

  =  ³⁵C₅ +  ³⁵C₄ + ³⁹C₄+ ³⁸C₄ + ³⁷C₄ + ³⁶C₄ 

  =  ³⁶C₅ + ³⁶C₄ + ³⁹C₄+ ³⁸C₄ + ³⁷C₄ 

  =  ³⁷C₅ + ³⁷C₄ + ³⁹C₄+ ³⁸C₄ 

=  ³⁸C₅ + ³⁸C₄  + ³⁹C₄ 

=  ³⁹C₅  + ³⁹C₄ 

=  ⁴⁰C₅ 

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