- Answers to Exercises 257
sincen−n 1 −···−nk− 1 −nk=0.
3.2.2.(a)n! (distribute positions instead of presents). (b)n(n−1)···(n−
k+ 1) (distribute as “presents” the firstkpositions at the competition and
n−kcertificates of participation). (c)
(n
n 1)
. (d) Chess seating in Diane’s
sense (distribute players to boards).
3.2.3. (a) [n= 8] 8!. (b) 8!·
( 8
4)
. (c) (8!)^2.
3.3 Anagrams
3.3.1. 13!/ 23.
3.3.2. COMBINATORICS.
3.3.3. Most: any word with 13 different letters; least: any word with 13
identical letters.
3.3.4. (a) 26^6.
(b)
( 26
4)
ways to select the four letters that occur; for each selection,( 4
2)
ways to select the two letters that occur twice; for each selection, we dis-
tribute 6 positions to these letters (2 of them get 2 positions); this gives
6!
2!2!ways. Thus we get
( 26
4)( 4
2)6!
2!2!. (There are many other ways to arrive
at the same number!)
(c) Number of ways to partition 6 into a sum of positive integers:
6=6=5+1=4+2=4+1+1=3+3=3+2+1=3+1+1+1=2+2+2=2+2+1+1=2+1+1+1+1=1+1+1+1+1+1,which makes 11 possibilities.
(d) This is too difficult in this form. What we meant is the following: how
many words of lengthnare there such that none is an anagram of another?
This means distributing( npennies to 26 children, and so the answer is
n+25
25
)
.
3.4 Distributing Money
3.4.1.
(
n−k− 1
k− 1)
.
3.4.2.
(
n+k− 1
+k− 1)
.
3.4.3.
(
kp+k− 1
k− 1)
.
3.5 Pascal’s Triangle
3.5.1. This is the same as
(n
k)
=
(n
n−k