I'm doing a initial course in combinatorics and all it went well until i've got stuck in this problem:
"7 people puts 7 presents on a table before a party. After the party every one of them takes only 1 present randomly. In how many ways can 3 (just 3) of them take their own present and the others take other presents that are not of them?".
I approached it by supposing a (1,2,3,4,5,6,7) list where every number is in their correct position. I tried many ways but i can't get to the answer, which is 315 (I did a program to count the correct permutations). A correct permutation may be: 6534217. Where 3,4 and 7 are in the correct places. May I be wrong in my supposition? I thought into substracting total (7!) with 7!/4! but i need more. Thanks to all and sorry by posting this low level question but I'm really stuck and I can't find a related problem or maybe I am not interpretating it the correct way.
After $3$ people get their own present, here are $!4=9$ ways to have $4$ people not get their own present. See derangements.
And, of course, there are ${7 \choose 3}=35$ ways to pick the $3$ people that get their own present.
Total: $35 \cdot 9 = 315$