Question

Ten cards, the 2 through 6 of spades and the 2 through the 6 of diamonds, are shuffled thoroughly and then taken one by one from the top of the deck and placed on the table.

I know how to solve the first question (I think), but how would I solve the second question? (below)

How many arrangements are possible?
10!=10x9x8x7x6x5xx4x3x2x1

What is the probability that each card is next to a card bearing the same numeral?

Answer 1

Whew. when I first read the problem I thought you were trying to find the probability that ANY two pairs were together. But, as I reread the question, you are looking for the probability that ALL pairs are together; much easier.

Choose the first pair to lay down, there are 5 possible pairs, within that pair there are 2 possibilities Spade/diamond or diamond/spade. So (5*2). For the second pair, you have 4 possibilities, again with 2 orders. And so on. So, the number of ways is (5*2)*(4*2)*(3*2)*(2*2)*(1*2)