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Permutations and Combinations Worksheet - Page 7

Permutations and Combinations Worksheet
  • Page 7
 61.  
In how many ways can you choose 2 even numbers and 3 odd numbers from the numbers 1, 2, 3, 4, 5, 6, 7?
a.
36
b.
28
c.
14
d.
12


Solution:

2, 4 and 6 are the even numbers, which are 3 in number.

1, 3, 5 and 7 are the odd numbers, which are 4 in number.

List all the possible combinations of choosing 2 even numbers as follows:
(2, 4), (2, 6)
(4, 6)

List all the possible combinations of choosing 3 odd numbers as follows:
(1, 3, 5), (1, 3, 7)
(3, 5, 7), (1, 5, 7)

List all the possible combinations of choosing 2 even numbers and 3 odd numbers as follows:
(2, 4, 1, 3, 5), (2, 4, 1, 3, 7), (2, 4, 3, 5, 7), (2, 4, 1, 5, 7)
(2, 6, 1, 3, 5), (2, 6, 1, 3, 7), (2, 6, 3, 5, 7), (2, 6, 1, 5, 7)
(4, 6, 1, 3, 5), (4, 6, 1, 3, 7), (4, 6, 3, 5, 7), (4, 6, 1, 5, 7)

So, 2 even numbers and 3 odd numbers can be chosen in 12 different ways.


Correct answer : (4)
 62.  
In how many ways can you arrange the letters in the word RANDOM?
a.
320
b.
720
c.
420
d.
520


Solution:

Number of possible outcomes for choosing the first letter = 6
[There are 6 letters in the word.]

Number of possible outcomes for choosing the second letter = 5
[There are 5 letters remaining to choose.]

Number of possible outcomes for choosing the third letter = 4
[There are 4 letters remaining to choose after choosing 2 letters.]

Number of possible outcomes for choosing the fourth letter = 3
[There are 3 letters remaining to choose after choosing 3 letters.]

Number of possible outcomes for choosing the fifth letter = 2
[There are 2 letters remaining to choose after choosing 4 letters.]

Number of possible outcomes for choosing the sixth letter = 1
[There is only 1 letter remaining to choose after choosing 5 letters.]

1stletter   2ndletter   3rdletter   4thletter   5thletter   6thletter
6×5×4×3×2×1=  720

[Multiply the number of possible outcomes of choosing each letter in the word.]

The letters in the word can be arranged in 720 ways.


Correct answer : (2)
 63.  
In how many ways can 8 employees be selected from 80 employees to form a committee?
a.
80P88!
b.
80P880!
c.
80P8
d.
None of the above


Solution:

Total number of employees = 80

Number of employees to be selected = 8

Number of ways to select 8 employees from 80 employees = 80C8

80C8 = 1 / 8! × 80P8 = 80P880!

There are 80P88! ways to select 8 employees from 80 employees.


Correct answer : (1)
 64.  
Find the number of three letter words that can be formed with the letters of the word TOP.
a.
9
b.
6
c.
3
d.
12


Solution:

The possible arrangements of the letters in the word TOP can be organized as shown below:
TOP, TPO, OPT, OTP, PTO and POT

So, the number of words that can be formed with the letters of the word TOP is 6.


Correct answer : (2)
 65.  
John went to a restaurant. The menu chart contains a list of 7 dishes. He wanted to order 5 dishes from the menu. State whether the way of ordering the dishes is a permutation or a combination.
a.
Combination
b.
Permutation


Solution:

A permutation is an arrangement in which the order is important and a combination is an arrangement in which the order does not matter.

The sequence of ordering dishes is not important here and different sequence of orders of same set of dishes form the same combination.

So, the ordering of dishes is a combination.


Correct answer : (1)
 66.  
Victor has to answer 2 questions out of 4 questions in an examination. State whether his selection of 2 questions is a permutation or a combination.
a.
Combination
b.
Permutation


Solution:

A permutation is an arrangement in which order is important and a combination is an arrangement in which order does not matter.

The sequence of selecting 2 questions is not important here and different sequence of orders of same set of questions form the same selection.

So, the selection of questions is a combination.


Correct answer : (1)
 67.  
State whether the arrangement of 3-lettered words from the alphabets C, H, E, M, I, S, T, R, Y is a permutation or a combination.
a.
Permutation
b.
Combination


Solution:

A permutation is an arrangement in which order is important and a combination is an arrangement in which order does not matter.

The ordering of alphabets is important here and different orders of same letters form different words.

So, the arrangement of 3-lettered words from the alphabets is a permutation.


Correct answer : (1)
 68.  
What is the value of 3P6?
a.
3
b.
6
c.
18
d.
Not defined


Solution:

The expression nPr can be defined as, the number of all permutations of n different things taken r at a time.

The value of r must be less than or equal to the value of n.

In 3P6, the value of r = 6 is greater than the value of n = 3.

The value of 3P6 is not defined.


Correct answer : (4)
 69.  
How many outcomes are possible, when three coins are tossed?

a.
6
b.
10
c.
3
d.
8


Solution:

Total number of coins = 3

Each coin has the outcome of either a Head (H) or Tail (T).

The possible outcomes of tossing three coins is HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

There are 8 possible outcomes when the three coins are tossed.


Correct answer : (4)
 70.  
John plans to study, watch TV and play football on a Sunday. In how many ways can he choose and arrange the activities?
a.
2
b.
6
c.
3
d.
1


Solution:

Total number of activities = 3

First activity can be chosen in 3 ways; second activity can be chosen in 2 ways and third activity can be chosen in 1 way.

By counting principle,
1st Activity   2nd Activity   3rd Activity
       3        ×         2        ×         1    =     6

John can choose and arrange the 3 activities in 6 different ways.


Correct answer : (2)

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