Suppose that S be the set of all the ordered 4-tuples (x, y, z, w) of the +ve integers, which are the solutions of x + y + z + w = 21. One such ordered tuple of solution is selected at random from S. Then find the probability that x > y.
Text Solution
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Sol. Total no. of +ve integral solutions of x + y + z + w = 21 is 21–1 C 4–1 = 1140
Let n be the no. of solutions in which x > y. then n be the solutions in which x < y and m be the solutions in which x = y. we must have 2n + m = 1140. Now, if x = y. then the equation is 2x + z + w = 21
If x = 1, z + w = 19 has 18 solutions
If x = 2, z + w = 17 has 16 solutions
If x = 9, z + w = 3 has 2 solutions
∴ m = 18 + 16 + ...... + 2 = 2 × 
⇒ 2n + 90 = 1140 ⇒ n = 525
Desired probability = 
= 
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