Question

The letters of the word ‘MADHURI’ are arranged in all possible ways. The number of arrangements in which there are 2 letters between R and D is

A. 360

B.480

C.960

D.720

Answer 1

Hint: At first we'll make cases to make the fixed position of R and D then will find the number of arrangements of the other letters and the letters R and D.
After finding the number of arrangements in all the cases, their sum will be the total number of required arrangements.

Complete step-by-step answer:
Given data: the word whose letters is to be arranged i.e. ‘MADHURI’
The number of ways of arranging n elements=n!
Here we have 7 letters in the word MADHURI
Let R and D are in 1st and 4th position
Now 5 places are left to arrange A, H, U, M, I
Therefore, the number of arrangements$ = 2 \times 5!$
A multiple of 2 is there as R and D can also interchange their position
The same will happen when R and D will be in (2nd and 5th position), (3rd and 6th position) and (4th and 7th position) i.e. the number of arrangements in each case will be$2 \times 5!$
Therefore we can say that the total number of arrangements of letters of the word ‘MADHURI’ such that there are 2 letters between R and D is $4 \times 2 \times 5!$
Total number of required arrangements$ = 4 \times 2 \times 5!$
$ = 8 \times 120$
$ = 960$
The letters of the word ‘MADHURI’ are arranged in all possible ways. The number of arrangements in which there are 2 letters between R and D is 960.
Option(3) is correct.

Note: An alternative method for the above problem can be
Since there should be 2 letters between R and D
Therefore selecting any 2 letters out of the remaining 5 and then arranging them $ = {}^5{C_2} \times 2!$
Number of ways of arranging R and D $ = 2!$
Now assuming R, D, and the 2 letters between them as one and then arranging the remaining 3 letters and this new combination$ = 4!$
Therefore the total number of arrangements$ = {}^5{C_2} \times 2! \times 2! \times 4!$
$ = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} \times 2! \times 2! \times 4 \times 3!$
$ = 2! \times 4 \times 5!$
$ = 8 \times 120$
$ = 960$, which is the same result as above