a2zinterviews

1.	In how many different ways can the letters of the word ‘JUDGE’ be arranged in such a way that the vowels always come together?
	A.  48
	B.  120
	C.  124
	D.  160

Solution

The word JUDGE has 5 different letters. When, the vowels UE are always together, they can be supposed to form one letter. Then, we have to arrange the letters JDG(UE). Now, 4 letters can be arranged in 4! = 24 ways. The vowels (UE) can be arranged in 2!= 2 ways. Required number of ways = (24 x 2) = 48.

2.	How many 4 letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
	A.  40
	B.  400
	C.  5040
	D.  2520

Solution

'LOGARITHMS' contains 10 different letters. Required number of words = Number of arrangements of 10 letters, taking 4 at a time. ‹=› 10p4 ‹=›(10 x 9 x 8 x 7) ‹=› 5040.

3.	In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
	A.  266
	B.  5040
	C.  11760
	D.  86400

Solution

Required number of ways ‹=› (8C5 ×10C6) ‹=›(8C3 ×10C4) =(8x7x6/3x2x1 ×10x9x8x7/4x3x2x1) ‹=›11760

4.	In how many ways can the letters of the word 'LEADER' be arranged?
	A.  72
	B.  144
	C.  360
	D.  720

Solution

The word 'LEADER' contains 6 letters namely 1L , 2 E, 1A,1D, and 1R. Required number of ways = 6! / (1!) (2!) (1!) (2!) = 360.

5.	How many words can be formed by using all letters of the word 'BIHAR'?
	A.  60
	B.  120
	C.  150
	D.  180

Solution

The word 'BIHAR' contains 5 different letters. Required number of words = 5P5 = 5! = (5 x 4x 3 x 2 x 1) = 120.