Permutations Practice Questions
2. Express 40  39 38 using factorials.
1. State the value of P(15, 3).
3.
Evaluate
.
4. State the number of words that can be formed by rearranging the letters of the word PROBABILITY.
5. State the number of ways that a committee of a president, treasurer, and secretary can be selected from a
students' council with 15 members.
6.
State the number of ways that the 9 members of the debating club can be lined up for a picture if Frasier must
be on the far left and Samantha and Charlotte must be together.
7. A bag contains 3 green blocks, 5 purple blocks, and 6 red blocks. If four blocks are drawn one at a time,
without replacement, determine the probability that the order is red, red, purple, green.
8. Three cards are drawn from a regular deck, one at a time, without replacement. Determine the probability that
the order is a jack, queen, and then a number 4, 5, or 6.
9. Solve for
if
.
10. The letters of the word INFINITY are scrambled. Determine the probability that F is the first letter and Y is
the last letter.
11. Three die are rolled simultaneously. Determine the probability that a sum of 10 will be rolled.
12. Four identical chocolate bars, three identical bags of chips, and 6 identical popsicles are lined up on a prize
table. Determine the number of ways that this could be done.
13. A coin is flipped, a die rolled, and a card drawn from a regular deck. Determine the probability that heads is
flipped, a five is rolled, and an ace drawn.
14. Two marbles are drawn from a bag containing 3 blue marbles and 7 orange marbles, one at a time, without
replacement. As well, two cards are removed from a regular deck, one at a time, without replacement.
Determine the probability that a blue marble and then an orange marble are drawn, as well as a club and then
a spade.
15 Determine the number of ways a party of 7 persons can arrange themselves around a circular table.
16. If 11 people are lined up and Ralf must be first, and Sonja, Sandra, and Kent must be together, explain in
words why the number of permutations is 241 920.
17. Two groups of 16 marbles are to be lined up. The first group has 16 marbles, all of a different colour, and the
second group has 6 red marbles, 2 blue marbles, and 8 pink marbles. Explain how the number of permutations
differs in these two situations, making reference to factorial notation.
18. Five digit numbers are made using the numbers 0 to 8. Determine the number of even numbers that can be
made if each digit can only be used once.
19. The letters of the word BALLAD are put into a hat and three are drawn out, one at a time, without
replacement. Determine the probability that two As are drawn in a row.
20. Solve for n if P(n, 3) = 6P(n – 1, 3).
21.
Solve for n if P(n, 2) = 72.
22. Six students are asked to secretly choose a number from 1 to 15. Determine the probability that at least two
students choose the same number to the nearest thousandth.
23. Determine the number of divisors for the number 3780, using the fact that a divisor of a number is the
multiplication of one or more of the factors of the number.
24. The 16 members of the Visual Arts council are seated for a presentation. Determine the probability, in
decimal form, that Julie and Tamar will not be seated next to each other.
25. A Canadian postal code consists of 6 characters of 3 letters alternating with 3 digits. An example is M4N
0R3. Mo would like to have a postal code that has his name somewhere in the postal code, and 7, his favourite
number. Determine the probability of this occurring.
Permutations Practice Questions
Answer Section
1. ANS:
The value of P(15, 3) is 2730.
2. ANS:
Using factorials, 40  39  38 is
.
3. ANS:
When
is evaluated, it equals 3 921 225.
4. ANS:
The number of words that can be formed is 9 979 200.
5. ANS:
The number of ways that a committee of a president, treasurer, and secretary can be chosen is 2730.
6. ANS:
The number of ways is 10 080.
7. ANS:
The probability of the order red, red, purple, green is
.
8. ANS:
The probability that the order is jack, queen, then 4, 5, or 6 is
.
9. ANS:
n = 42
10. ANS:
The probability that F is the first letter and Y the last is
.
11. ANS:
The probability that a sum of 10 will be rolled is
.
12. ANS:
The number of ways that this could be done is 60 060.
13. ANS:
The probability is
.
14. ANS:
The probability is
.
15. ANS:
The number of ways is 720.
PROBLEM
16. ANS:
Without any restrictions, the number of permutations is 11!, but the first position never changes and so the
number of ways becomes 10!. Sonja, Sandra. and Kent can be treated as one person because they must always
be together, so the number of permutations becomes 8!. The three of them can however be permuted 3! ways,
and so the number of permutations of the people is (8!)(3!) = 241 920.
17. ANS:
With the first group, the number of permutations is 16! because the number of choices for the first position is
16, the number of choices for the second position is 15, and so on. The second group, however, contains
marbles with the same colour, so if two marbles of the same colour are put in two particular positions,
switching these positions does not create a new permutation. For the six red marbles, which can be permuted
6! different ways, each of these ways produces the same permutation, so you must divide 16! by 6!. Thus, all
of these same permutations are only counted as one. The same effect is caused with the blue and pink
marbles, so you must divide by 2! and 8! also. Therefore, the number of permutations of the second group is
16! divided by 6!, 2!, and 8!.
18. ANS:
There are two cases. For numbers that end with zero, the number of possibilities is (8)(7)(6)(5)(1) = 1680. For
numbers that do not end with zero, the number of possibilities is (7)(7)(6)(5)(4) = 5880. Therefore, the total
number of possibilities is 1680 + 5880 = 7560.
19. ANS:
The total number of ways that the numbers can be drawn is the sum of three possibilities. First of all, the
number of permutations with 2 As is (3 other letters)(3 positions) = 9.
Secondly, the number of permutations with 2 Ls also equals 9.
If 3 different letters are used, the number of permutations is (4)(3)(2) = 24.
Therefore, the total number of permutations is 9 + 9 + 24 = 42. The number of possible permutations with 2
As in a row is (3 other letters)(2 positions) = 6. Therefore the probability of drawing 2 As in a row is
.
20. ANS:
21. ANS:
Since cannot be negative,
.
22. ANS:
The probability is P(at least two choose the same number) = 1 – P(no two choose the same number).
23. ANS:
3780 = (2)(2)(3)(3)(3)(5)(7). A factor of 3780 will be composed of zero, one or two 2s, zero, one, two, or
three 3s, zero or one 5 and zero or one 7. Therefore, there are 3 ways to choose 2s, 4 ways to choose 3s, 2
ways to choose 5 and 2 ways to choose 7s. A divisor must have at least one factor though, so it cannot have
zero, 2s, 3s, 5s, and 7s. Therefore, the number of divisors will be (3)(4)(2)(2) – 1 (the case of zero, 2s, 3s, 5s,
and 7s) = 47.
24. ANS:
The probability is P(Julie and Tamar will not be next to each other)
25. ANS:
Number of ways that MO and 7 could be in the postal code = (3 positions for M and O) 
(26 letters)  (3 positions for 7)  (10 numbers)  (10 numbers) = 23 400.
Number of possible codes
Therefore, P(MO and 7 in the code)