Question 1187855
<pre>
b)How many different ways are there to get exactly 1 pair (i.e. 2 cards with the
same rank)?

To get one pair, we first:

Choose the 1 rank.  That's 16 ranks choose 1, 16C1=16 ways.

Then we choose the 2 suits.  That's 9 suits choose 2, 9C2=36 ways.

We choose the ranks of the other three cards different from the 1 rank that the
pair has, and also different from each other, so there will not be another pair:
That's 15 other ranks choose 3, 15C3 = 455 ways.

We choose the suit for the card of the other 3 with the lowest rank 9 ways.
We choose the suit for the card of the other 3 with the middle rank 9 ways.
We choose the suit for the card of the other 3 with the highest rank 9 ways.

That's (16C1)(9C2)(9)(9)(9) = (16)(36)(9)(9)(9) = 419904

So, the number of ways of getting exactly 1 pair is 419904 ways. 

The number of ways to get any 5 cards is 144C5 = 481008528

So the probability is 419904 out of 481008528 or 419904/481008528 which
reduces to 2916/3340337 or 0.0008730, to 7 decimal places.

Edwin</pre>