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Examples


Example
Evaluate
  1. 10C4
  2. 12C7
  3. 6C6
Solution
  1. 10C4 =
  2. 12C7 =
  3. 6C6 =
Example
  1. If 2nCr = 2nCr+2, find r.
Solution
  1. We shall use the result nCr = nCs r = s or r + s = n
    2nCr = 2nCr+2
    Thus we can conclude as follows
    ⇒ r = r + 2 or r + (r + 2) = 2n
    But r r + 2, therefore 2r + 2 = 2n r = n - 1.
  1. If nC10 = nC14 find nC20 and 25Cn.
Solution
  1. nC10 = nC14
10 + 14 = n or n = 24.
Thus nC20 = 24C20 =
          = 23 × 22 × 21 = 10626
and 25Cn = 25C24 = 25C1 = 25.
  1. If 18Cr = 18Cr+2, find rC5
Solution
  1. 18Cr = 18Cr+2 r + r + 2 = 18 or r = 8.
Thus rC5 = 8C5 =
  1. If n+2C8: n-2P4 = 57 : 16 find n.
Solution
  1. n+2C8 : n-2P4= 57 : 16


⇒ (n+2) (n+1) n (n-1) = (57) (2520)
⇒ (n2 + n - 2) (n2 + n) = 14360 (1)
Put n2 + n - 1 = y, so that (1) can be written as (y - 1) (y + 1) = 143640
⇒ y2 - 1 = 143640 y2 = 143641 = 3792 y = 379             [Q y > 0]
⇒ n2 + n - 1 = 379 n2 + n - 380 = 0                   [putting the value of y]
  .
As n cannot be negative, n = 19.
  1. If nPr = nPr+1 and nCr = nCr-1 ; find n and r.
Solution
  1. We are given nPr = nPr+1

Next, nCr = nCr-1                                                                 [given]


n - r + 1 = r
n - 2r = -1                                                                 (2)
From (1) and (2), n = 3, r = 2.




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