# Exercise 1 (Ross 1.7).

```Solutions to Problem Set 2.1
Due: Friday September 3, 2004
Exercise 1 (Ross 1.7).
(a) In how many ways can 3 boys and 3 girls sit in a row?
(b) In how many ways can 3 boys and 3 girls sit in a row if the boys and girls are each to
sit together?
(c) In how many ways if only the boys must sit together?
(d) In how many ways if no two people of the same sex are allowed to sit together?
Solution.
(a) There are 6! = 720 ways 6 people can sit in a row.
(b) We can have either BBBGGG or GGGBBB, so disregarding the order within boys and
within girls, there are 2 possibilites. There are 3! ways to order the boys, and 3! ways
to order the girls, so all together there are (2)(3!)(3!) = 72 ways to do this.
(c) If the boys are regards as a unit, then we can first order the objects {B, G1 , G2 , G3 },
where B is the unit of the three boys, and G1 , G2 , G3 are the girls. There are 4! ways
to do this. Then the boys can be ordered in 3! ways. Thus the answer is 4!3! = 144.
(d) We can have either BGBGBG or GBGBGB. For each such arrangement, there are 3!
ways to arrange the boys, and 3! ways to arrange the girls, so the answer is (2)(3!)(3!) =
72.
Exercise 2 (Ross 1.9). A child has 12 blocks, of which 6 are black, 4 are red, 1 is white,
and 1 is blue. If the child puts the blocks in a line, how many arrangements are possible?
Solution. There are 12 spots to be filled: 6 of them are black, 4 are red, 1 is white, and 1
is blue. Think of the spots as a collection which we are to partition into 4 subcollections of
sizes 6,4,1, and 1. In other words, we must decide for each spot whether it is to be black,
red, white, or blue, but we must do this in such a way so that there are 6 blacks, 4 reds, 1
white, and 1 blue. There are
12
12!
=
= 27720
6, 4, 1, 1
6!4!1!1!
1
ways to do this.
Another way to think about this as follows. Suppose the blocks are labelled, so we have
α1 , α2 , . . . , α6 , β1 , β2 , . . . , β4 , γ, δ, where the α’s are the black blocks, the β’s are the red
blocks, γ is the single white block, and δ is the single blue block. Then all the blocks are
distinguishable from one another and there are 12! such arrangements. But for each such
arrangements, there are 6!4!1!1! arrangements which differ only in the subscripts. By this,
I mean that I can permute the α’s amongst themselves and the β’s amongst themselves in
6!4! ways, and this will not alter the arrangements of colors, only of the subscripts. Thus,
considering color alone, there are 12!/(6!4!1!1!) distinguishable arrangements.
Exercise 3 (Ross 1.13). Consider a group of 20 people. If everyone shakes hands with
everyone else, how many handshakes take place?
Solution. How many ways to pick two people from 20? There are 20
such pairs, so the
2
answer is (20)(19)/2 = 190.
Exercise 4 (Ross 1.15). A dance class consists of 22 students, 10 women and 12 men. If
5 men and 5 women are to be chosen and then paired off, how many results are possible?
12
Solution. First, we can pick 5 men and 5 women in 10
ways. There are 5! ways to
5
5
pair up the chosen 5 mean with the 5 women. [Think of lining up the men in a row. There
are 5! ways of then lining up the women to create the pairs.] Thus the answer is
10 12
5! .
5
5
Exercise 5 (Ross 1.19). From a group of 8 women and 6 men a committee consisting of
3 men and 3 women is to be formed. How many different committees are possible if
(a) 2 of the men refuse to serve together.
(b) 2 of the women refuse to serve together.
(c) 1 man and 1 woman refuse to serve together.
Solution.
(a) Distinguish one of the men in thepair which refuses to serve with one another, and
suppose he is “Bob”. There are 42 ways to choose the men in the committee so that
Bob is included, as there are 4 men who he would serve with, of which we have to
2
choose 2. If Bob is not in the committee, then there are 53 ways to pick the men.
[There are 5 other men besides Bob, of which we must choose 3.] There are 83 ways
to choose the women on the commitee. Thus the number of committees is
4
5
8
+
= 896 .
2
3
3
(b) The same reasoning gives
6
7
6
+
= 1000 .
2
3
3
5 7
Call the man
in
the
feuding
couple
Bob
again.
There
are
committees containing
2 3
5 8
Bob, and 3 3 committess not containing Bob. Thus the total number of committees is
5 7
5 8
+
= 910 .
2 3
3 3
Exercise 6 (Ross 1.25). The game of bridge is played by 4 players, each of whom is dealt
13 cards. How many bridge hands are possible?
Solution. There are
52
52!
=
13, 13, 13, 13
(13!)3
ways to split the cards into 4 groups, each of size 13. This does not distinguish between
the order of cards within a hand.
3
```