by using derangement with fix point (combinatoric)

Question

Suppose you want to permute the first five positive integers. How many ways are there if (a) you require the third integer to be in its natural position but the others are not. (b) you require the third integer to be in its natural position and exactly one other integer in its natural position as well, but the others are not. 

Answer 1

The first 5 positive integers are assumed to be 1, 2, 3, 4, 5.

Fixing 3 leaves 24 ways of arranging the numbers 1, 2, 4, 5:

12345 12354 14325 14352 15324 15342

21345 21354 24315 24351 25314 25341

41325 41352 42315 42351 45312 45321

51324 51342 52314 52341 54312 54321

(a) From this list we can eliminate all those where the integers are in their natural position, leaving the 3 where it is. 21354 24351 25314 41352 45312 45321 51324 54312 54321 (9 arrangements).

(b) Now we want one other integer than 3 in its natural position:

14352 15324 24315 25341 41325 42315 51342 52314 (8 arrangements); 2 with 1 and 3; 2 with 2 and 3, 2 with 3 and 4; 2 with 3 and 5, making 8 in all.