1) In what ways the letters of the word "RUMOUR" can be arranged?

1. 180
2. 150
3. 200
4. 230

The word RUMOUR consists of 6 words in which R and U are repeated twice.
Therefore, the required number of permutations = Or, = 180

Hence, 180 words can be formed by arranging the word RUMOUR.

2) In what ways the letters of the word "PUZZLE" can be arranged to form the different new words so that the vowels always come together?

1. 280
2. 450
3. 630
4. 120

The word PUZZLE has 6 different letters.

As per the question, the vowels should always come together.
Now, let the vowels UE as a single entity.
Therefore, the number of letters is 5 (PZZL = 4 + UE = 1)
Since the total number of letters = 4+1 = 5
So the arrangement would be in 5P5 = = = 5! = 5*4*3*2*1 = 120 ways.

#### Note: we know that 0! = 1

Now, the vowels UE can be arranged in 2 different ways, i.e., 2P2 = 2! = 2*1 = 2 ways

Hence, the new words, which can be formed after rearranging the letters = 120 *2 = 240

As we known z is occurring twice in the word ‘PUZZLE’ so we will divide the 240 by 2.

So, the no. of permutation will be = 240/2 = 120

3) In what ways can a group of 6 boys and 2 girls be made out of the total of 7 boys and 3 girls?

1. 50
2. 120
3. 21
4. 20

We know that nCr = nC(n-r)

The combination of 6 boys out of 7 and 2 girls out of 3 can be represented as 7C6 + 3C2
Therefore, the required number of ways = 7C6 * 3C2 = 7C(7-6) * 3C(3-2) = = 21

Hence, in 21 ways the group of 6 boys and 2 girls can be made.

4) Out of a group of 7 boys and 6 girls, five boys are selected to form a team so that at least 3 boys are there on the team. In how many ways can it be done?

1. 645
2. 734
3. 756
4. 612

We may have 5 men only, 4 men and 1 woman, and 3 men and 2 women in the committee.

So, the combination will be

as we know that

nCr= So, (7C3 * 6C2) + (7C4 * 6C1) + (7C5)
Or, + + Or, 525 +210+21 = 756

So, there are 756 ways to form a committee.

5) A box contains 2 red balls, 3 black balls, and 4 white balls. Find the number of ways by which 3 balls can be drawn from the box in which at least 1 black ball should be present.

1. 64
2. 48
3. 32
4. 96

The possible combination could be (1 black ball and 2 non-black balls), (2 black balls and 1 non- black ball), and (only 3 black balls).

Therefore the required number of combinations = (3C1 * 6C2) + (3C2 * 6C1) + (3C3)
r, + + = 45+18+1 = 64

Permutation and Combination Test Paper 2
Permutation and Combination Concepts

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