# Compute number of combinations

I got really stuck with this task:

Nine of ten cards, among which there is an ace of hearts, are distributed to three players so that the first one receives 3, the second - 4, and the third - 2 cards. How many cards combinations exist, where an ace of hearts gets to a third player?

I think that number of combinations formula is $$c_{9}^{1} * c_{3}^{1} * c_{2}^{1} = 6$$

Am I right? I will be so grateful for your help!)

• What does your $c_a^b$ mean? – Parcly Taxel Mar 12 at 13:24
• As well as that, are all the cards distinct? – Parcly Taxel Mar 12 at 13:27
• @ParclyTaxel could it be OP means $$c_n^k = \binom{n}{k}?$$ – gt6989b Mar 12 at 13:31
• The ace of hearts is among those nine cards or among those ten cards? I think among those ten, but just to be sure? – TStancek Mar 12 at 13:36
• @TStancek of course, among them – Lord of Programs Mar 12 at 14:15

Just give the ace of hearts to the third player first. Then allocate the rest of the hands:

• The third player gets a second card to complete their hand in $$\binom91=9$$ ways, leaving $$8$$ cards
• The second player gets their hand in $$\binom84=70$$ ways, leaving $$4$$ cards
• The first player gets their hand in $$\binom43=4$$ ways

Thus there are $$9\times70\times4=2520$$ ways to distribute the cards.

• Is it solution for combinations, where an ace of hearts gets to a third player? Because it looks like all ways to distribute the cards among 3 players – Lord of Programs Mar 12 at 14:13
• @LordofPrograms Yes, combinations. – Parcly Taxel Mar 12 at 14:17
• Cool, thanks a lot) – Lord of Programs Mar 12 at 15:42