My question is this : how many distinct two digit numbers can be produced from numbers $4, 3, 3, 1$? When applying the formula $$\frac{4!}{(4-2)!2!}$$ you come up with $6$, yet when doing the problem manually, I come up with $7$ numbers, namely $41, 43, 14, 13, 33, 31, 34$. Can anybody explain this discrepancy?
Thanks.
Your formula $\frac {4!}{(4-2)!2!}=6$ counts the number of two element combinations from four distinct objects. For a two digit number, you would want two element permutations, as order matters. This would be $\frac {4!}{(4-2)!}=12$ Then you have to account for the fact that two of your objects (the $3$'s) match. For a problem this small, you can do (and have done) a hand count. Otherwise you need to break it into cases. There are two numbers that do not involve a 3, four that involve a single 3, and one that has two 3's, for a total of 7.
You're correct, its 7. Here's why. The permutation finds distinct two digit numbers. Your set has two threes in it. By the way, your permutation should've been $12=4*3$
