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Question

The value of x satisfying the relation 11(xC3)=24((x+1)C2) is
  1. 9
  2. 8
  3. 11
  4. 12
  5. 10

A
12
B
10
C
8
D
9
E
11
Solution
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11(xC3)=24(x+1C2)
11×x!(x3)!3!=24×(x+1)!(x1)!2!
11×x(x1)(x2)(x3)!(x3)!3! =24×(x+1)(x)(x1)!(x1)!2!
11(x1)(x2)=72(x+1)
Therefore, x=10 satisfy the above equation.

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