30.3 PERMUTATIONS AND COMBINATIONS
the number ofdistinguishablepermutations is only
n!
n 1 !n 2 !···nm!, (30.32)since theith group of identical objects can be rearranged inni! ways without
changing the distinguishable permutation.
A set of snooker balls consists of a white, a yellow, a green, a brown, a blue, a pink, a
black and 15 reds. How many distinguishable permutations of the balls are there?In total there are 22 balls, the 15 reds being indistinguishable. Thus from (30.32) the
number of distinguishable permutations is
22!
(1!)(1!)(1!)(1!)(1!)(1!)(1!)(15!)=
22!
15!
= 859 541 760.
30.3.2 CombinationsWe now consider the number ofcombinationsof various objects when their order
is immaterial. Assuming all the objects to be distinguishable, from (30.31) we see
that the number of permutations ofkobjects chosen fromnisnPk=n!/(n−k)!.
Now, since we are no longer concerned with the order of the chosen objects, which
can be internally arranged ink! different ways, the number of combinations ofk
objects fromnis
n!
(n−k)!k!≡nCk≡(
n
k)
for 0≤k≤n, (30.33)where, as noted in chapter 1,nCkis called thebinomial coefficientsince it also
appears in the binomial expansion for positive integern, namely
(a+b)n=∑nk=0nC
kakbn−k. (30.34)A hand of 13 playing cards is dealt from a well-shuffled pack of 52. What is the probability
that the hand contains two aces?Since the order of the cards in the hand is immaterial, the total number of distinct hands
is simply equal to the number of combinations of 13 objects drawn from 52, i.e.^52 C 13.
However, the number of hands containing two aces is equal to the number of ways,^4 C 2 ,
in which the two aces can be drawn from the four available, multiplied by the number of
ways,^48 C 11 , in which the remaining 11 cards in the hand can be drawn from the 48 cards
that are not aces. Thus the required probability is given by
(^4) C 248 C 11
(^52) C 13 =