Is it something like (5 choose 3) * 5! * 3! ?
I'd like to confirm this for an exam tomorrow.
My train of thought was 'Choose 3 from 5, to be doubled up on, and then arrange the first 5 matches, and finally to arrange the doubles.'
I'd like to confirm this for an exam tomorrow.
My train of thought was 'Choose 3 from 5, to be doubled up on, and then arrange the first 5 matches, and finally to arrange the doubles.'
-
We count this by inclusion-exclusion.
#(total number of mappings) - #(missing at least one element of [5]) + #(missing at least 2 elements of [5]) - #(missing at least 3 elements of [5]) + #(missing at least 4 elements of [5]) - #(missing all 5 elements of [5])
This simplifies to
5^8 - C(5, 1) * (5 - 1)^8 + C(5, 2) * (5 - 2)^8 - C(5, 3) * (5 - 3)^8 + C(5, 4) * (5 - 4)^8 - 0
= 5^8 - C(5, 1) * 4^8 + C(5, 2) * 3^8 - C(5, 3) * 2^8 + C(5, 4) * 1^8.
I hope this helps!
#(total number of mappings) - #(missing at least one element of [5]) + #(missing at least 2 elements of [5]) - #(missing at least 3 elements of [5]) + #(missing at least 4 elements of [5]) - #(missing all 5 elements of [5])
This simplifies to
5^8 - C(5, 1) * (5 - 1)^8 + C(5, 2) * (5 - 2)^8 - C(5, 3) * (5 - 3)^8 + C(5, 4) * (5 - 4)^8 - 0
= 5^8 - C(5, 1) * 4^8 + C(5, 2) * 3^8 - C(5, 3) * 2^8 + C(5, 4) * 1^8.
I hope this helps!
-
Pardon the delay; as you see, this problem is a bit subtle and easy to mess up. Posting a wrong answer on this could have been most embarrassing on my part!
Report Abuse