Determine the number of derangements of (2, 4, 6, 1, 3, 5) that end with integer 2, 4 and 6 in some order?(a) 128(b) 29(c) 54(d) 36The question was asked in semester exam.This interesting question is from Counting topic in section Counting of Discrete Mathematics

Determine the number of derangements of (2, 4, 6, 1, 3, 5) that end wi

1.	Determine the number of derangements of (2, 4, 6, 1, 3, 5) that end with integer 2, 4 and 6 in some order?(a) 128(b) 29(c) 54(d) 36The question was asked in semester exam.This interesting question is from Counting topic in section Counting of Discrete Mathematics
Answer» Correct option is (d) 36 Explanation: The PLACE of 2, 4, 6 is SPECIFIED i.e. each of them will GET their place out of the last 3 places only. So 1, 3, 5 will automatically get one of the places in the first 3 places. This MUST ensure that 2, 4 and 6 occupies one of the last 3 places each and 1, 3 and 5 one of 1st 3 places each. Hence, 1, 3 and 5 can be arranged in 3! ways and 2, 4 and 6 also in 3! Ways. So, no of such derangements = 3! * 3! = 6 * 6 = 36.

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