I am having trouble with this problem: The six digits 1,1,1,2,3,3 are permuted. How many different permutations begin with the digit 2? I am assuming I can reverse 1 and 3 several times?
Thanks
Hello Koeppel1The answer is 10. Once the 2 has been fixed in the first position, you have to arrange the remaining 5 items. If they are all different, this can be done in 5! ways. Because of the repetition of the 1's you need to divide this by 3!, since that's the number of ways in which the 1's can be re-arranged among themselves. Then you also divide by 2!, because of the duplication of the 3's.
(I'm not sure what you mean by 'reverse 1 and 3 several times'?)
Answer: $\displaystyle \frac{5!}{3!2!}= 10$
Grandad