How do you solve for number of variations for a specific number of repetitions?

Asked Oct 20 '20 at 22:18

Active Oct 20 '20 at 22:18

Viewed 12 times

I am trying to determine how many different permutations I can make with 4 values {1,2,3,4} for an 8 chain sequence. Normally, I would simply do 4 to the power of 8. However, there can only be two repetitions of each value -- so 11223344 is a possibility, but 11111111 is not.

How would I go about solving for the number of possible permutations? I couldn't find a formula for this online, and I'm having trouble figuring it out.

asked Oct 20 '20 at 22:18

Zofia Korsak

This is a multinomial coefficient, $\binom{8}{2,2,2,2}=\dfrac{8!}{2!2!2!2!}=\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$ – JMoravitz Oct 20 '20 at 22:24
thank you! But why is that the case? – Zofia Korsak Oct 20 '20 at 22:35
2

Choose the positions occupied by 1s, then choose positions occupied by 2s and so on... alternatively directly use the combinatorial definition of multinomial coefficients – JMoravitz Oct 20 '20 at 22:37
I understand now! Thank you! – Zofia Korsak Oct 20 '20 at 22:42

How do you solve for number of variations for a specific number of repetitions?

0 Answers0