Distributing half a deck of cards 2 cards each are dealt to 5 persons from a half a deck of cards (with just 2 suits).
How many hands are there in which none of the 5 get a face card pair (KK, QQ or JJ), and what is the probability of such a hand ?
I am finding it difficult to count because of numerous branches.  
 A: Use inclusion/exclusion.  Let
$$\eqalign{
  {\cal U}&=\{\hbox{all possible deals}\}\cr
  K&=\{\hbox{deals in which someone gets two kings}\}\cr
  Q&=\{\hbox{deals in which someone gets two queens}\}\cr
  J&=\{\hbox{deals in which someone gets two jacks}\}\ .\cr}$$
Then the number of hands you want is
$$|\overline K\cap\overline Q\cap\overline J|=|{\cal U}|-|K|-|Q|-|J|
  +|K\cap Q|+|K\cap J|+|Q\cap J|-|K\cap Q\cap J|\ .$$
I'll calculate some of these and leave the rest for you.
To count $\cal U$,


*

*choose two cards from $26$ for the first person. . . . . $C(26,2)$ ways

*choose two cards from $24$ for the second person. . . . . $C(24,2)$ ways

*and so on.


Hence
$$|{\cal U}|=C(26,2)C(24,2)C(22,2)C(20,2)C(18,2)\ .$$
To count $K\cap Q$,


*

*choose the person who receives two kings. . . . . $5$ ways

*choose the person who receives two queens. . . . . $4$ ways

*choose cards for the others as above.


Hence
$$|K\cap Q|=5\times4\times C(22,2)C(20,2)C(18,2)\ .$$
See how you go with the rest.
