A bag contains 50 tickets numbered 1, 2, 3, . . ., 50 of which five are drawn at random and arranged in ascending order of magnitude x1
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a
20C2 50C5
b
2C2 50C5
c
20C2×29C2 50C5
d
None of these
answer is C.
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Detailed Solution
Five tickets out of 50 can drawn in 50C5 ways. Since x130, should come from 20 tickets numbered 31 to 50 in 20C2 ways. So, favourable number of cases =29C229C2 Hence, required probability = 29C2×29C2 50C5