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MR. A randomly picks 3 distinct numbers from the set ${1, 2, 3, ...., 9}$ and arranges them in descending order to form a three digit number. MR. B randomly picks 3 distinct numbers from the set ${1, 2, 3, ...., 8}$ and also arranges then in descending order to form 3 digit numbers.

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Probability that MR. A’s 3 digit number is always greater than MR. B’s 3 digit number is $\frac{1}{3}$ .

Probability that MR. A and MR. B has the same 3 digit number is $\frac{1}{84}$ .

Probability that MR. A’s 3 digit number is greater than MR. B’s 3 digit number is $\frac{37}{56}$ .

Probability that MR. A’s 3 digit number is greater than MR. B’s 3 digit number is $\frac{1}{3}$ .

answer is A, B, C.

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Detailed Solution

When 9 is not included $= \frac{^{8} C_{3} .^{8} C_{3} -^{8} C_{3}}{2}$
And required probability is $= \frac{1}{3} + = \frac{^{8} C_{3} .^{8} C_{3} -^{8} C_{3}}{{2.}^{9} C_{3} .^{8} C_{3}} = \frac{37}{56}$

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