Using permutations and combinations for counting (solutions)

1. V63.0233: Theory of Probability Solutions
Worksheet for Sections 1.3–1.4 : Permutations and Combinations June 30, 2009
1. Among the 16 applicants for four different teaching positions in an elementary school, only
ten have master’s degrees.
(i) In how many ways can these positions be filled?
(ii) In how many ways can these positions be filled with applicants having master’s degrees?
(iii) If one of the positions requires a master’s degree, while for the others it’s optional, in how
many ways can the four positions be filled?
Solution. (i) Notice that the positions are described as different so order matters. We have 16
ways to fill the first spot, 15 to fill the second, etc., until you fill four spots:
16 × 15 × 14 × 13 = 43, 680
(ii) We have 10 ways to fill the first, spot, 9 ways to fill the second, and so on:
10 × 9 × 8 × 7 = 5040
(iii)
2. How many anagrams of the word MASSACHUSETTS can you find?
Solution. Of the 13 letters, we have two A’s, four S’s, and two T’s. This gives a total of
13!
= 64, 864, 800
4!2!2!
anagrams.
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2. 3. A student takes a true-false test of 15 questions. In how many different ways can he or she
mark this test and get
(i) three right and 12 wrong?
(ii) six right and nine wrong?
(iii) 12 right and three wrong?
Solution. (i) We can choose the incorrect three to be any of the questions numbered 1 through
15. Hence there are ( )
15
= 455
3
different ways to do this.
(ii) ( )
15
= 5005
6
(iii) ( )
15
= 455
12
(Notice there are exactly the same number of ways to get three right as there are to get three
wrong.)
4. Rework the MASSACHUSETTS problem by choosing the slots for each of the letters. For
instance, there are 13 positions for the M. Of the 12 remaining, we need to choose two for the Ts.
Of the 10 remaining, we need to choose four for the Ss, etc.
Solution. Following the directions, we can choose ) of 13 positions for the M. The 2 As need to
( any
12
go in any of the remaining 12 slots; there are ways to do this. And so on. We get
2
( )( )( )( )( )( )( )( )
13 12 10 6 5 4 3 2
= 64, 864, 800
1 2 4 1 1 1 1 2
M A S C H U E T
This is the same thing we get in Problem 2.
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