Instructions: This Monk test contains JK Bank Permutation and Combination Problems with answer. This will help you in JK bank exam preparation.
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Directions to Solve
In the following the questions choose the word which best expresses the meaning of the given word.
Results
#1. 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?
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  = (^{7}C_{3} x ^{6}C_{2}) (^{7}C_{4} x ^{6}C_{1}) (^{7}C_{5})  




= (525 210 21)  
= 756. 
#2. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
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.
#3. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
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
in  5!  = 20 ways. 
3! 
Required number of ways = (2520 x 20) = 50400.
#4. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Answer: Option C
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (^{7}C_{3} x ^{4}C_{2})  


= 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 
= 5! 
= 5 x 4 x 3 x 2 x 1  
= 120. 
Required number of ways = (210 x 120) = 25200.
#5. In how many ways can the letters of the word 'LEADER' be arranged?
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!) 
#6. 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?
Answer: Option D
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways 
= (^{6}C_{1} x ^{4}C_{3}) (^{6}C_{2} x ^{4}C_{2}) (^{6}C_{3} x ^{4}C_{1}) (^{6}C_{4})  
= (^{6}C_{1} x ^{4}C_{1}) (^{6}C_{2} x ^{4}C_{2}) (^{6}C_{3} x ^{4}C_{1}) (^{6}C_{2})  


= (24 90 80 15)  
= 209. 
#7. How many 3digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Answer: Option D
Explanation:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
#8. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Answer: Option C
Explanation:
Required number of ways  = (^{8}C_{5} x ^{10}C_{6})  
= (^{8}C_{3} x ^{10}C_{4})  


= 11760. 
#9. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
Answer: Option C
Explanation:
We may have(1 black and 2 nonblack) or (2 black and 1 nonblack) or (3 black).
Required number of ways  = (^{3}C_{1} x ^{6}C_{2}) (^{3}C_{2} x ^{6}C_{1}) (^{3}C_{3})  


= (45 18 1)  
= 64. 
#10. In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
Answer: Option C
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = ^{3}P_{3} = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = ^{3}P_{3} = 3! = 6.
Total number of ways = (6 x 6) = 36.
#11. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Answer: Option A
Explanation:
Required number of ways = (^{7}C_{5} x ^{3}C_{2}) = (^{7}C_{2} x ^{3}C_{1}) =  7 x 6  x 3  = 63.  
2 x 1 
#12. How many 4letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
Answer: Option C
Explanation:
‘LOGARITHMS’ contains 10 different letters.
Required number of words  = Number of arrangements of 10 letters, taking 4 at a time. 
= ^{10}P_{4}  
= (10 x 9 x 8 x 7)  
= 5040. 
#13. In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
Answer: Option C
Explanation:
In 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.
#14. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
Answer: Option B
Explanation:
The word ‘OPTICAL’ contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
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