Answer
Verified
91.2k+ views
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If the distance between 1st crest and the third crest class 11 physics JEE_Main
A man of mass 50kg is standing on a 100kg plank kept class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main