How many permutations p of (1, 2, 3, 4, 5, 6, 7) satisfy all of the…
Question
How many permutations p of (1, 2, 3, 4, 5, 6, 7) satisfy all of the following: the values in positions 1, 2, 3 are each even (i.e. p(1), p(2), p(3) are all chosen from {2,4,6}), and the permutation has no fixed point at all (p(i) is not equal to i for every i from 1 to 7)?
Step-by-step solution
The even values available are exactly {2,4,6} (three of them), and positions 1,2,3 must receive all three even values in some order.
So positions 1,2,3 hold a permutation of {2,4,6}, and positions 4,5,6,7 hold a permutation of the remaining values {1,3,5,7}.
Arrangements of {2,4,6} in positions 1,2,3 with p(2)!=2: total 3!=6, minus those with 2 in position 2 (2!=2), giving 4.
Values are 1,3,5,7; the forbidden matches are p(5)!=5 and p(7)!=7 (4 and 6 are not in the value set, so positions 4 and 6 are automatically fine).
So we need permutations of (1,3,5,7) into positions (4,5,6,7) avoiding value 5 in position 5 and value 7 in position 7.
Total arrangements 4! = 24. Subtract those with 5 in pos5 (3!=6) or 7 in pos7 (3!=6), add back both (2!=2): 24 - 6 - 6 + 2 = 14.
Final answer56
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