# Ways to select a hand of 9 cards from a deck of 36

This is a very basic self learning question, the scenario is there are 36 cards of 4 suits from 1 to 9 of each suit. One can pick a hand of 9 cards. My question is how many ways can someone pick a hand which contains all the 1's given that:

1) the order of the cards in the hand doesn't matter. Initially I thought this would be 36! /(4! X 5! X 27!) - from the concept of arranging among the 36 cards that 4 always chosen 1's, 5 selected and 27 non-selected out of the 36, but it turns out to be combination(32,5) as 4 1's are always chosen and we are left to choose the rest 5 from the 32 cards. I get this explanation but what am I doing wrong with my arranging approach stated formerly?

2) the order of the cards in the hand matter. Kind of lost on this one.

For part (a), you tried: $$\dfrac {36! }{4! \times 5! \times 27!} = \dbinom{36}{4, 5, 27}$$

This multinomial coefficient is the ways to pick 9 cards out of 36 into two separate hands, one of 4 cards, and one of 5 (and 27 remain in the deck).

$$\dfrac {36! }{4! \times 5! \times 27!} =\underbrace{\dfrac{36!}{4!\times 32!}}_{\text{choose 4 of 36}}\times\underbrace{\dfrac{32!}{5!\times 27!}}_{\text{choose 5 of 32}}$$

What you wanted however, was the ways to choose 4 of the 4 '1' cards and 5 of the 32 not '1' cards into the same hand.

Thus: $$\dfrac{4!}{4!\times 0!}\!\!\times\!\!\dfrac{32!}{5!\times 27!} = \dbinom{4}{4}\dbinom{32}{5}$$

For part (b), when the order of the cards in the hand matters, each possible combination of $9$ cards has $9!$ permutations.

So that would be $$\dfrac{32!\times 9!}{5!\times 27!}$$

You need to use the binominals

$$\binom{n}{k}=\frac{n!}{(n-k)!\times k!}$$

$$\binom{36}{9}=\frac{36!}{27!\times 9!}=94143280$$