Conditional Permutations with allowed repeating

Question

Let's say we have four elements A, B, C, and D.

How can I calculate number of all possible permutations (order matters) for set of 4, if repetitions are allowed, but it is required to have at least one B and one C element in each set?

How can I calculate similar task for same elements, same conditions, but for set of 5?

Hint: inclusion/exclusion ideas work here – Mark Bennet May 06 '20 at 20:15 — Mark Bennet, May 06 '20 at 20:15

score 3 · Accepted Answer · answered May 06 '20 at 20:21

3

The total orderings are $4^4$, since each blank has $4$ choices. Now calculate the orderings which have no $B’s$ or no C’s:

$$n(\bar B \cup \bar C) = n(\bar B) + n(\bar C) - n(\bar B \cap \bar C) \\ =3^4 + 3^4 -2^4 \\ =146 $$

Hence, our desired answer is $4^4 -146 =110$

You can use a similar procedure for any number of blanks.

answered May 06 '20 at 20:21

Vishu

14,469

Thank you very much. Very logical approach. Can't ask for more. – Ωmega May 06 '20 at 20:24

Conditional Permutations with allowed repeating

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