Question

In how many ways can the letters of the word ARRANGE be arranged?

Answer 1

Hint: In this question, we have to find the ways in which the letters of the word ARRANGE can be arranged. We should note that while considering the number of ways, we should treat the same letter present at different places to be identical. Thus, as A and R are present two times, we should consider the A and R present at the different positions to be identical. As there are seven letters in total no of ways of arrangement should be equal to the ratio of 7! and the product of 2! and 2! as A and R are present two times in the given word.

Complete step-by-step answer:
The given word is ARRANGE. Thus, any rearrangement should consist of 7 letters
To find the total number of ways, we should multiply the number of ways in which the first place can be filled and the number of ways in which the second place can be filled after the first place is filled and so on up to the 7th place…………………………………….(1.1)
Now, if we consider each letter to be different, the first place can be filled by any of the letters and thus in 7 ways. After the first place has been filled, there are 6 letters remaining and thus the second place can be filled in 6 ways. Similarly, the 3rd place can be filled in 5 ways and so on up to the 7th place which can be filled in just one way.
Thus, the no. of arrangements considering each letter to be different is $ 7\times 6\times 5\times 4\times 3\times 2\times 1 $ ………(1.1)
However, in ARRANGE, the two A’s and the two R’s are identical letters, thus two arrangements in which only the R’s are interchanged will not result in different arrangements. For example, if we put the first R in ARRANGE at the fifth place and second R of ARRANGE at the seventh place, and if we put the first R in ARRANGE at the seventh place and second R of ARRANGE at the fifth place keeping positions of all other letters unchanged, they would give the same word.
The formula for arranging n letters in which $ {{p}_{2}} $ letters are identical of the first kind, $ {{p}_{2}} $ letters are identical of the first kind and so on upto $ {{p}_{r}} $ letters of the rth kind is given by
 $ \dfrac{n!}{{{p}_{1}}!{{p}_{2}}!...{{p}_{r}}!}...........................(1.2) $
Thus, as here there are 7 letters are total and 2 letters are of the type R and 2 letters are of the type A, using equation (1.2), we get
Total arrangements of the word ARRANGE= $ \dfrac{7!}{2!2!}=\dfrac{7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 1\times 2\times 1}=1260...............(1.3) $
Thus 1260 is the required answer.

Note: We should note that in equation (1.3), we divided 2! And 2! As there were two A and two R in ARRANGE. If there would have been n letters of the same type then we should have divided by n!, for example, if there were 3 A’s then we should have divided by 3! and so on.