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Solution. Consider any one of the (n + m)! possible orderings, and note that any permutation

of the red balls among themselves and of the blue balls among themselves

does not change the sequence of colors. As a result, every ordering of colorings corresponds

to n! m! different orderings of the n + m balls, so every ordering of the

colors has probability n!m!

(n+m)! of occurring.

For example, suppose that there are 2 red balls, numbered r1, r2, and 2 blue balls,

numbered b1, b2. Then, of the 4! possible orderings, there will be 2! 2! orderings that

result in any specified color combination. For instance, the following orderings result

in the successive balls alternating in color, with a red ball first:

r1, b1, r2, b2 r1, b2, r2, b1 r2, b1, r1, b2 r2, b2, r1, b1

Therefore, each of the possible orderings of the colors has probability 4


= 16


occurring. .


A poker hand consists of 5 cards. If the cards have distinct consecutive values and

are not all of the same suit, we say that the hand is a straight. For instance, a hand

consisting of the five of spades, six of spades, seven of spades, eight of spades, and

nine of hearts is a straight. What is the probability that one is dealt a straight?

Solution. We start by assuming that all



possible poker hands are equally

likely. To determine the number of outcomes that are straights, let us first determine

the number of possible outcomes for which the poker hand consists of an ace, two,

three, four, and five (the suits being irrelevant). Since the ace can be any 1 of the 4

possible aces, and similarly for the two, three, four, and five, it follows that there are

45 outcomes leading to exactly one ace, two, three, four, and five. Hence, since in 4 of

these outcomes all the cards will be of the same suit (such a hand is called a straight

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