Answer
Verified
40.2k+ 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
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
Other Pages
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
Which of the following is a buffer solution A CH3COOH class 11 chemistry JEE_Main