Question for Abhiram Rao 5

Total number of words that can be formed from ABHIRAM = $\dfrac{7!}{2!} = 2520$ (because 7 letters can be arranged in 7! ways and two A's repeat which can be arranged in 2! ways).

Total number of words that end with M = ?
ABHIRAM is of seven alphabets. The last alphabet can only have one letter in it and that is M. The remaining six alphabets can be arranged in $\dfrac{6!}{2!} = 360$ ways (because six letters can be arranged in 6! ways and two A's that repeat can be arranged in 2! ways)

So, total number of words that do not end with M = $2520 - 360 = \boxed{2160}$ .
$\boxed{\text{Q.E.D}}$

• We can visualize this problem by boxes representing possible positions for M, hence last position is left empty without box.

$\text{\_A\_A\_B\_H\_I\_R}$

Number of positions M can be = 6C1 = 6 (6 empty boxes for 1 letter)

Number of Combinations the remaining AABHIR letters = $\frac{6!}{2!}$ = (where 2! to compensate for the 2 "A" s)

Hence, 6*360 = 2160

First We calculate the total number of words which can be formed with $\color{#3D99F6}{A}\color{#D61F06}{B}\color{#CEBB00}{H}\color{cyan}{I}\color{#69047E}{R}\color{#EC7300}{A}\color{#20A900}{M}$ , which is equal to $\dfrac{7!}{2!}$

Now let's find the number of words which end with $\color{#20A900}{M}$ ,which is $\dfrac{6!}{2!}$

Our Answer is $\dfrac{7!}{2!}-\dfrac{6!}{2!} \implies 6!\times3 =\color{#3D99F6}{\boxed{2160}}$

We have 6 choices for last letter excluding M. Then we have 6!/2! ways to arrange other letters as we can include M among them and there are two A's. Then answer is 6*6!/2! = 2160