How to Find the Value of n and r in Combination :
In this section, we will learn, how to find the values of n and r in the given combination.
Question 1 :
If nC12 = nC9 find 21Cn.
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
21Cn = 21C21 = 1
Hence the answer is 1.
Question 2 :
If 15C2r−1 = 15C2r+4, find r.
Solution :
If nCx = nCy ==> 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 nPr = 720, and nCr = 120, find n, r.
Solution :
nPr = 720
n!/(n - r)! = 720 ---(1)
nCr = 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 :
nCr + n Cr−1 = n+1Cr
= 16C4 + 16C5
= 17C5 ---> R.H.S
Question 5 :
Prove that 35C5 + ∑ 4r=0 (39−r)C4 = 40C5.
Solution :
L.H.S
= 35C5 + ∑ 4r=0 (39−r)C4
= 35C5 + 39C4+ 38C4 + 37C4 + 36C4 + 35C4
By using the property :
nCr + n Cr−1 = n+1Cr
= 35C5 + 35C4 + 39C4+ 38C4 + 37C4 + 36C4
= 36C5 + 36C4 + 39C4+ 38C4 + 37C4
= 37C5 + 37C4 + 39C4+ 38C4
= 38C5 + 38C4 + 39C4
= 39C5 + 39C4
= 40C5
After having gone through the stuff given above, we hope that the students would have understood, how to find the values of n and r in the given combinations.
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