In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
A.
360
B.
480
C.
720
D.
5040
Answer: Option C
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
2.
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
A.
564
B.
645
C.
735
D.
756
Answer: Option D
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways
= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
=
7 x 6 x 5
x
6 x 5
+ (7C3 x 6C1) + (7C2)
3 x 2 x 1
2 x 1
= 525 +
7 x 6 x 5
x 6
+
7 x 6
3 x 2 x 1
2 x 1
= (525 + 210 + 21)
= 756.
3.
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
A.
810
B.
1440
C.
2880
D.
50400
Answer: Option D
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =
7!
= 2520.
2!
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
Required number of ways = (2520 x 20) = 50400.
4.
In how many ways can the letters of the word 'LEADER' be arranged?
A.
72
B.
144
C.
360
D.
720
Answer: Option C
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways =
6!
= 360.
(1!)(2!)(1!)(1!)(1!)
5.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A.
210
B.
1050
C.
25200
D.
21400
Answer: Option C
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (7C3 x 4C2)
=
7 x 6 x 5
x
4 x 3
3 x 2 x 1
2 x 1
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves