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A Learning Portal from Recruitment India
In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together? 
10080
4989600
120960
None of these
Answer with explanation
Answer: Option CExplanation
n the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters =  8!  = 10080. 
(2!)(2!) 
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters =  4!  = 12. 
2! 
Required number of words = (10080 x 12) = 120960.
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In how many different ways can the letters of the word ‘LEADING’ be arranged such that the vowels should always come together?
720
420
122
120
Answer with explanation
Answer: Option AExplanation
The word ‘LEADING’ has 7 letters. It has the vowels ‘E’,’A’,’I’ in it and these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. that is, LDNG(EAI).
Hence we can assume total letters as 5 and all these letters are different. Number of ways to arrange these letters
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120
In the 3 vowels (EAI), all the vowels are different. Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6
Hence, required number of ways
=120×6=720
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In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
266
5040
11760
86400
Answer with explanation
Answer: Option CExplanation
Required number of ways  = (^{8}C_{5} x ^{10}C_{6})  
= (^{8}C_{3} x ^{10}C_{4})  


= 11760. 
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In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected so that at least one boy should be selected?
159
209
201
212
Answer with explanation
Answer: Option BExplanation
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In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
810
1440
2880
50400
Answer with explanation
Answer: Option DExplanation
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
in  5!  = 20 ways. 
3! 
Required number of ways = (2520 x 20) = 50400.
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In how many ways can 5 men and 2 ladies be arranged at a round table if two ladies are never together?
720
420
580
480
Answer with explanation
Answer: Option DExplanation
total no. of arrangements = 6! = 720
No. of arrangements in which 2 ladies are together = 2 X 5! = 240
And the arrangements in which two ladies are never together = 720 – 240 = 480
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In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
209
205
194
159
Answer with explanation
Answer: Option AExplanation
Any permutation and combination question which consists of “at least” word can be answered in 2 ways:
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A box contains 4 red, 3 white, and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls in different colors
168
12
24
48
Answer with explanation
Answer: Option CExplanation
This question seems to be a bit typical, isn’t, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways
Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24
Please note that we have multiplied the combination results, we use to add when there is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.
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How many 3 digit numbers are divisible by 4?
125
158
225
252
Answer with explanation
Answer: Option CExplanation
A number is divisible by 4 when its last two digits are divisible by 4
For this, the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96
By the formula, a_{n} = a + (n1)d
====>>>> 96 = 0 + (n1)*4
==== >>>> n = 25
so there are 25 choices for last 2 digits and 9 choices (19) for the 1st digit
====>>>> so total 9*25 = 225
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Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.
310
320
330
340
Answer with explanation
Answer: Option CExplanation
Here the order of choosing the elements doesn’t matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
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