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Combinations Worksheet

Combinations Worksheet
  • Page 1
 1.  
(99) = ?
a.
9!
b.
9
c.
1


Solution:

(99) = 9!(9 - 9)! · 9!
[(nr) = nCr = n!(n - r)!  r!.]

= 9!9! · 0!

= 1
[0! = 1.]


Correct answer : (3)
 2.  
A box contains 5 color pencils. How many different sets of 3 pencils can be chosen from it?
a.
10
b.
4
c.
6
d.
20


Solution:

The number of ways of choosing 3 sets from 5 is 5C3.

= 5!(5 - 3)! · 3!
[nCr = n!(n - r)!  r!.]

= 10

10 sets of 3 pencils can be formed.


Correct answer : (1)
 3.  
Find the number of ways of selecting 2 vowels and 2 consonants from the letters of the set {I, N, T, E, R, V, A, L}.
a.
21
b.
60
c.
13
d.
30


Solution:

The number of vowels in the set = 3
[The vowels in the set are I, E, A.]

The number of consonants = 5
[The consonants in the set are N, T, R, V, L.]

2 vowels can be selected from the 3 vowels in 3C2 ways.

2 consonants can be selected from the 5 consonents in 5C2 ways.

The number of ways of selecting 2 vowels and 2 consonants from the letters of the given set = 3C2 · 5C2
[Use the fundamental counting principle.]

= 3!(3 - 2)! · 2! · 5!(5 - 2)! · 2!
[Use nCr = n!(n - r)!  r!.]

= 3 · 2!1! · 2! · 5 · 4 · 3!3! · 2! = 30
[Simplify.]

The number of ways of selecting 2 vowels and 2 consonants from the letters of the given set = 30


Correct answer : (4)
 4.  
How many 4 card hands that contain exactly 3 aces and 1 queen can be chosen from a 52 card deck?
a.
12
b.
4
c.
16
d.
8


Solution:

There are 4 aces and 4 queens in the 52-card deck. 3 aces can be chosen from 4 in 4C3 ways, and 1 queen can be chosen from 4 in 4C1 ways.

So, the number of 4 card hands that contain exactly 3 aces and 1 queen is (4C3)(4C1).

= 4!(4 - 3)! · 3! · 4!(4 - 1)! · 1!
[nCr = n!(n - r)!  r!.]

= 4 · 3!1! · 3! · 4 · 3!3! · 1! = 16
[1! = 1.]


Correct answer : (3)
 5.  
If nC2 = 28, then what is the value of n?
a.
8
b.
7, - 8
c.
- 7
d.
8, - 7


Solution:

nC2 = 28

n!(n - 2)!  2! = 28
[nCr = n!(n - r)!  r!.]

n(n - 1)(n Ã¢â‚¬â€œ 2)!(n - 2)!  2! = 28

n(n - 1) = 56

n2 – n - 56 = 0

(n - 8)(n + 7) = 0
[Factor.]

n = 8 or n = - 7

n cannot be negative. So, n = 8.


Correct answer : (1)
 6.  
Find the number of permutations of the elements in the set {F, O, U, R}.
a.
3
b.
4
c.
6
d.
24


Solution:

The number of elements in the set = 4

The number of permutations of 4 distinct elements of the set is 4P4

= 4! = 24
[nPn = n!.]


Correct answer : (4)
 7.  
Find the number of permutations of the elements in the set {17, 19, 21, 23, 25, 27}.
a.
25
b.
720
c.
162
d.
150


Solution:

Number of elements in the set = 6

Number of permutations of 6 distinct elements of the set is 6P6.

= 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
[nPn = n!.]


Correct answer : (2)
 8.  
Find the number of 8 letter permutations from the letters of the word KEYBOARD.
a.
40320
b.
1
c.
8
d.
64


Solution:

The number of letters of the word is 8.
[Write the number of letters of the word.]

The number of 8 letter permutations from the 8 distinct letters of the word is 8P8 = 8! = 40320
[nPn = n!.]

The number of 8 letter permutations from the 8 distinct letters of the word KEYBOARD is 40320.


Correct answer : (1)
 9.  
In how many ways can 2 different books be arranged on a shelf?
a.
2!
b.
3!
c.
5!
d.
4!


Solution:

The number of ways of arranging 2 different books on a shelf is 2P2

= 2!
[nPn = n!.]


Correct answer : (1)
 10.  
How many different 2 digit numbers can be formed using {2, 3, 4, 5, 6, 7, 8}, if repetition of a digit is not allowed?
a.
120
b.
42
c.
72
d.
5040


Solution:

The number of permutations of 7 digits taken 2 at a time is: 7P2 = 7!(7 Ã¢â‚¬â€œ 2)!
[nPr = n!(n - r)!.]

= 7!5! = 42


Correct answer : (2)

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