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Quiz: Permutations and Combinations

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Number of Questions: 40

Question: 26 -

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Options:

4320

2160

120

720

Answer:

720

Solution:

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.

Question: 27 -

How many four-digit numbers, each divisible by 4 can be formed using the digits 5, 6, 7, 8, 9, repetition of digits being allowed in any number?

Options:

150

125

100

Answer:

125

Solution:

For a number to be divisible by 4, the last two digits of the number have to be 56, 76, 96, 68 or 88. The first two digits of these numbers can be filled in 5 x 5 = 25 ways (numbers can be repeated). Hence, in total there are 25 x 5 = 125 such numbers.

Question: 28 -

In a class there are 20 boys and 25 girls. In how many ways can a boy and a girl be selected?

Options:

400

600

500

Answer:

500

Solution:

We can select one boy from 20 boys in 20 ways.
We select one girl from 25 girls in 25 ways
We select a boy and girl in 20 * 25 ways i.e., = 500 ways.

Question: 29 -

A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?

Options:

1023

511

Answer:

1023

Solution:

Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.

Hence, the required number of ways = 4⁵ - 1.

= 2¹⁰ - 1 = 1024 - 1 = 1023

Question: 30 -

A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included in a particular student?

Options:

Answer:

10

Solution:

If students are A, B, C, D, E and F; we can have ⁶C₃ groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get ⁵C₂= 10 ways of doing it.

Quiz: Permutations and Combinations

Solution: For a number to be divisible by 4, the last two digits of the number have to be 56, 76, 96, 68 or 88. The first two digits of these numbers can be filled in 5 x 5 = 25 ways (numbers can be repeated). Hence, in total there are 25 x 5 = 125 such numbers.

Solution: We can select one boy from 20 boys in 20 ways.We select one girl from 25 girls in 25 waysWe select a boy and girl in 20 * 25 ways i.e., = 500 ways.

Solution: Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted. Hence, the required number of ways = 45 - 1. = 210 - 1 = 1024 - 1 = 1023

Solution: If students are A, B, C, D, E and F; we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get 5C2 = 10 ways of doing it.

Solution:

For a number to be divisible by 4, the last two digits of the number have to be 56, 76, 96, 68 or 88. The first two digits of these numbers can be filled in 5 x 5 = 25 ways (numbers can be repeated). Hence, in total there are 25 x 5 = 125 such numbers.

Solution:

We can select one boy from 20 boys in 20 ways.
We select one girl from 25 girls in 25 ways
We select a boy and girl in 20 * 25 ways i.e., = 500 ways.

Solution:

Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.

Hence, the required number of ways = 4⁵ - 1.

= 2¹⁰ - 1 = 1024 - 1 = 1023

Solution:

If students are A, B, C, D, E and F; we can have ⁶C₃ groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get ⁵C₂= 10 ways of doing it.