Instructions: This Monk test contains JK Bank Permutation and Combination Problems with answer. This will help you in JK bank exam preparation.

This is a FREE online JK Bank Monk Test on topic JK Bank Permutation and Combination Problems with answer. You DO NOT pay money to anyone to attend this test.
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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

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#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 = (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.

#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)

= (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
= 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
= (6C1 x 4C3) (6C2 x 4C2) (6C3 x 4C1) (6C4)
= (6C1 x 4C1) (6C2 x 4C2) (6C3 x 4C1) (6C2)
= (6 x 4) 6 x 5 x 4 x 3 6 x 5 x 4 x 4 6 x 5
2 x 1 2 x 1 3 x 2 x 1 2 x 1
= (24 90 80 15)
= 209.

#7. How many 3-digit 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 = (8C5 x 10C6)
= (8C3 x 10C4)
8 x 7 x 6 x 10 x 9 x 8 x 7
3 x 2 x 1 4 x 3 x 2 x 1
= 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 non-black) or (2 black and 1 non-black) or (3 black).

 Required number of ways = (3C1 x 6C2) (3C2 x 6C1) (3C3)
= 3 x 6 x 5 3 x 2 x 6 1
2 x 1 2 x 1
= (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 = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 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 = (7C5 x 3C2) = (7C2 x 3C1) = 7 x 6 x 3 = 63.
2 x 1

#12. 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?

Answer: Option C

Explanation:

‘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.

#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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