2
$\begingroup$

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!)

$\endgroup$
  • $\begingroup$ What does your $c_a^b$ mean? $\endgroup$ – Parcly Taxel Mar 12 at 13:24
  • $\begingroup$ As well as that, are all the cards distinct? $\endgroup$ – Parcly Taxel Mar 12 at 13:27
  • $\begingroup$ @ParclyTaxel could it be OP means $$c_n^k = \binom{n}{k}?$$ $\endgroup$ – gt6989b Mar 12 at 13:31
  • $\begingroup$ The ace of hearts is among those nine cards or among those ten cards? I think among those ten, but just to be sure? $\endgroup$ – TStancek Mar 12 at 13:36
  • $\begingroup$ @TStancek of course, among them $\endgroup$ – Lord of Programs Mar 12 at 14:15
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Lord of Programs Mar 12 at 14:13
  • $\begingroup$ @LordofPrograms Yes, combinations. $\endgroup$ – Parcly Taxel Mar 12 at 14:17
  • $\begingroup$ Cool, thanks a lot) $\endgroup$ – Lord of Programs Mar 12 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.