Combinatorics Through Guided Discovery, 2004a

1.2. Basic Counting Principles 19
⇒ Problem 44. We first gave this problem as Problem 12.a. Now we have
several ways to approach the problem. A tennis club has 2n members. We
want to pair up the members by twos for singles matches.
(a) In how many ways may we pair up all the members of the club? Give
at least two solutions different from the one you gave in Problem 12.a.
(You may not have done Problem 12.a. In that case, see if you can find
three solutions.) (h)
(b) Suppose that in addition to specifying who plays whom, for each
pairing we say who serves first. Now in how many ways may we
specify our pairs? Try to find as many solutions as you can. (h)
• Problem 45. (This becomes especially relevant in Chapter 6, though it
makes an important point here.) In how many ways may we attach two
identical red beads and two identical blue beads to the corners of a square
(with one bead per corner) free to move around in (three-dimensional)
space? (h)
⇒
Problem 46. While the formula you proved in Problems 35 and 39.d is very
useful, it doesn’t give us a sense of how big the binomial coefficients are.
We can get a very rough idea, for example, of the size of ( 2n )
n
by recognizing
that we can write (2n) n /n! as 2n n · 2n−1 n+1
n−1 ···
1
, and each quotient is at least
2, so the product is at least 2 n . If this were an accurate estimate, it would
mean the fraction of n-element subsets of a 2n-element set would be about
2 n /2 2n = 1/2 n , which becomes very small as n becomes large. However it
is pretty clear the approximation will not be a very good one, because some
of the terms in that product are much larger than 2. In fact, if ( 2n )
k
were the
same for every k, then each would be the fraction 1
2n+1 of 22n . This is much
larger than the fraction 1
2
. But our intuition suggets that ( 2n ) n n
is much larger
than ( 2n 2n
)
1
and is likely larger than ( )
n−1
so we can be sure our approximation
is a bad one. For estimates like this, James Stirling developed
(
a formula to
√2πn )
approximate n! when n is large, namely n! is about n n /e n .Infact
the ratio of n! to this expression approaches 1 as n becomes infinite. a We
write this as
n! ∼ √ 2πn nn
e n .
We read this notation as n! is asymptotic to √ 2πn nn
e
. Use Stirling’s formula
to show that the fraction of subsets of size n in an 2n-element set is
n
approximately 1/ √ πn. This is a much bigger fraction than 1
2
! n