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 

Question 6 :

If ⁿ C ₁₂ = ⁿ C ₈, then n is equal to 

a) 20   b)  12   c) 6    d)  30

Solution :

ⁿ C ₁₂ = ⁿ C ₈ ----(1)

ⁿ C _r = ⁿ C _n-r

ⁿ C ₁₂ = ⁿ C _{n - 12}

From (1)

ⁿ C _{n - 12} = ⁿ C ₈

n - 12 = 8

n = 8 + 12

n = 20

So, the value of n is 20.

Question 7 :

Find r, If ¹⁵ C _r : ¹⁵ C _{r - 1} = 11 : 5

Solution :

¹⁵ C _r : ¹⁵ C _{r - 1} = 11 : 5

¹⁵ C _{15- r} : ¹⁵ C _{r - 1} = 11 : 5

¹⁵ C _{15- r} / ¹⁵ C _{r - 1} = 11 : 5

[15!/(15 - 15 + r)! (15 - r)!] / [15! / (15 - r + 1)! (r - 1)!] = 11/5

(16 - r)! (r - 1)!/r!(15 - r)! = 11/5

(16 - r)(15 - r)! (r - 1)!/r!(15 - r)! = 11/5

(16 - r) (r - 1)/r(r - 1)! = 11/5

(16 - r)/r = 11/5

Doing cross multiplication, we get

5(16 - r) = 11r

80 - 5r = 11r

80 = 11r + 5r

16r = 80

r = 80/16

r = 5

So, the value of r is 5.

Question 8 :

If ⁿP_r = 336, ⁿC_r = 56. Find n and r and hence find ^n–1C_r–1.

Solution :

ⁿP_r = 336, ⁿC_r = 56

n!/(n - r)! = 336 -----(1)

n!/(n - r)! r! = 56 -----(2)

[n!/(n - r)!][1/r!] = 56

336/r! = 56

r! = 336/56

r! = 6

r! = 3 x 2 x 1!

r = 3

Applying r = 3 in (2), we get

n!/(n - 3)! 3! = 56

n!/(n - 3)! = 56 (6)

n (n - 1)(n - 2)(n - 3)!/(n - 3)! = 56 (6)

n (n - 1)(n - 2) = 8 x 7 x 6

n = 8

^n–1C_r–1 = ^8–1C_3–1

= ⁷C₂

= 7!/(7 - 2)! 2!

= 7! / 5! 2!

= (7 x 6) x 5!/5! x 2

= (7 x 6)/2

= 7 x 3

= 21

Question 9 :

Find n if ²ⁿC₃ : ⁿC₃ = 11 : 1

Solution :

²ⁿC₃ : ⁿC₃ = 11 : 1

²ⁿC₃ / ⁿC₃ = 11 / 1

[(2n)! / (2n - 3)! 3!] / [n! /(n - 3)! 3!] = 11/1

[(2n)(2n - 1)(2n - 2)(2n - 3)! / (2n - 3)! 3!] x [(n - 3)! 3! / n!] = 11/1

[(2n)(2n - 1)(2n - 2)] x [(n - 3)!/n!] = 11/1

[(4n)(2n - 1)(n -1)] x [(n - 3)!/n(n - 1)(n - 2)(n - 3)!] = 11/1

4(2n - 1)/(n - 2) = 11/1

8n - 4 = 11(n - 2)

8n - 4 = 11n - 22

8n - 11n = -22 + 4

-3n = -18

n = 6

So, the value of n is 6.

Question 10 :

Show that (n + 2) n! = n! + (n + 1)!

Solution :

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

R.H.S :

= n! + (n + 1)!

= n! + (n + 1) n!

= n!(1 + n + 1)

= n! (n + 2)

= (n + 2) n!

L.H.S

So, it is proved.

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