The number of derangements of {1, 2, 3, ..., n} can be computed using the recursive formula D(n) = (n-1) [D(n-1) + D(n-2)], where D(1) = 0 and D(2) = 1.
Using this formula, we can compute the number of derangements of {1, 2, 3, 4, 5, 6, 7} that begin with the integers 1, 2, and 3 in some order as follows:
Let's assume that the first three integers in the derangement are 1, 2, and 3. We can consider two cases:
Case 1: 1 is moved to a position other than the second position. In this case, there are 4 possible positions for 1, and once 1 is placed in a position, there are (5-2)! = 3! = 6 ways to derange the remaining 5 elements. Therefore, the number of derangements in this case is 4 x 6 x D(4), where D(4) is the number of derangements of the remaining 4 elements.
Case 2: 1 is moved to the second position. In this case, there are 2 possible positions for 2 and once 2 is placed, there are (5-3)! = 2! = 2 ways to derange the remaining 5 elements. Therefore, the number of derangements in this case is 2 x 2 x D(5), where D(5) is the number of derangements of the remaining 5 elements.
Using the recursive formula, we can compute D(4) and D(5) as follows: D(4) = 3 [D(3) + D(2)] = 3[2+1] = 9 D(5) = 4 [D(4) + D(3)] = 4[9+2] = 44
Therefore, the total number of derangements of {1, 2, 3, 4, 5, 6, 7} that begin with the integers 1, 2, and 3 in some order is:
4 x 6 x D(4) + 2 x 2 x D(5) = 4 x 6 x 9 + 2 x 2 x 44 = 348.
Therefore, there are 348 derangements of {1, 2, 3, 4, 5, 6, 7} that begin with the integers 1, 2, and 3 in some order.
