6) Out of 7 constants and 4 vowels, how many words of 3 constants and 2 vowels can be formed?

1. 21020
2. 25200
3. 10500
4. 21400

We can combine 3 consonants and 2 vowels out of 7 consonants and 4 vowels in a way

7C3 * 4C2 = = 210

That means 210 groups having 3 consonants and 2 vowels.

Each group consists of 5 words that mean the possible arrangement of the letters is 5!

Or, 5! = 5*4*3*2*1 = 120 ways.

Therefore, the required number of words = 210*120 = 25200.

7) How much 4-digit number can be formed from the digits 2, 3, 4, 5, 6, and 7 which are divisible by 5 in such a way that digits should not repeat.

1. 25
2. 30
3. 35
4. 60

A number is divisible by 5 if the number ends with 0 or 5, but we don't have 0 in the given digits that means 5 should come at the unit place.

Now, one of the remaining 5 digits (2, 3, 4, 6, and 7) can come at the tens place.
Similarly, we can fill the hundreds place by one of the remaining 4 digits.
Therefore, the thousands place can be filled by one of the remaining 3 digits.
Hence, the required number of the numbers = 1*5*4*3 = 60.

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

1. 120
2. 240
3. 360
4. 480

The word CRICKET has 7 different letters, but ATQ, the vowels should always come together.
Now, let the vowels IE as a single entity.
Therefore, the number of letters is CRCKT = 5 in which C is repeated twice, and IE = 1
Since the total number of letters = 5+1 = 6
So the arrangement (permutation) would be = = = 5*4*3= 60 ways.

#### Note: Here 2! is taken in the denominator, because the letter C is repeated twice.

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

So, the new words that can be formed = 60*2 = 120.

9) In what ways the letters of the word ?MACHINE? can be arranged so that the vowels occupy only the odd positions?

1. 212
2. 326
3. 576
4. 400

The word machine consists of 7 letters in which there are 3 vowels and 4 consonants.
ATQ, the vowels A, I, E can be placed at any of the position out of 1, 3, 5, and 7.
That means the number of ways to arrange the vowels = 4P3 = = = 4*3*2*1 = 24

Similarly, the number of ways to arrange the consonants = 4P4 = = 4*3*2*1 = 24

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

Therefore the required numbers of ways = 24*24 = 576

10) In what ways the letters of the word ?ACTORS? can arrange so that the vowels occupy only the even positions?

1. 212
2. 144
3. 576
4. 400

The word ACTORS consist of 6 letters in which there are 2 vowels and 4 consonants
ATQ, the vowels A, O can be placed at any of the position out of 2, 4, and 6.
That means the number of ways to arrange the vowels = 3P2 = = = 3*2 = 6 ways

Similarly, the number of ways to arrange the consonants = 4P4 = = 4*3*2*1 = 24 ways

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

Therefore the required numbers of ways = 6*24 = 144

Permutation and Combination Test Paper 1
Permutation and Combination Concepts   