Questions & Answers

Question

Answers

a) 120,6

b) 720,12

c) 120,12

d) 720,24

Answer

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Hint: These types of questions involve the concept of permutation. First look at the letters we have got, see how many of them are the same and how many are unique so that we can have a fair idea about how many places we have to fill and then continue.

We have total 6 letters, out of which we have 3 S’s and 3 different letters i.e. ‘A’, ’I’ and ‘T’.

Since, we have 3 same letters, it would be easy to start with this

We can place all ‘S’s either at even places or at odd places i.e. we have 2 ways of placing ‘S’

∴ The remaining letters can be placed at the remaining places in 3! Ways i.e. 6 ways.

This is because if we fix ‘A’ at first place, then ‘I’ can have the rest of the two places, fixing ‘I’ at the second place, then ‘T’ is left with only one place which doesn’t need to be fixed. Same pattern follows if we fix ‘I’ in the first place. Thus we wrote 3!

∴ Total number of ways = 2×3!

=2×3×2×1

=12

Similarly, total no. of ways in which letter ‘S’ can be placed with other letters =5!

= 5×4×3×2×1

=120, 12

∴ Option ‘C’ is the right answer.

Note: Permutation of a set is an arrangement of its elements into a sequence or linear order, or if it is already ordered, a rearrangement of it’s elements. In this question also we rearranged the letters of the word ‘ASSIST’ which is acting as a set.

We have total 6 letters, out of which we have 3 S’s and 3 different letters i.e. ‘A’, ’I’ and ‘T’.

Since, we have 3 same letters, it would be easy to start with this

We can place all ‘S’s either at even places or at odd places i.e. we have 2 ways of placing ‘S’

∴ The remaining letters can be placed at the remaining places in 3! Ways i.e. 6 ways.

This is because if we fix ‘A’ at first place, then ‘I’ can have the rest of the two places, fixing ‘I’ at the second place, then ‘T’ is left with only one place which doesn’t need to be fixed. Same pattern follows if we fix ‘I’ in the first place. Thus we wrote 3!

∴ Total number of ways = 2×3!

=2×3×2×1

=12

Similarly, total no. of ways in which letter ‘S’ can be placed with other letters =5!

= 5×4×3×2×1

=120, 12

∴ Option ‘C’ is the right answer.

Note: Permutation of a set is an arrangement of its elements into a sequence or linear order, or if it is already ordered, a rearrangement of it’s elements. In this question also we rearranged the letters of the word ‘ASSIST’ which is acting as a set.