Number of arrangements of letter BHARAT taken $3$ at a time I have to find the number of arrangements of the letters of BHARAT taken $3$ at a time.
Now, according to me answer should be $P(6,3)$ and then divide by $2!$, but it ain't that way in textbook. May I know what's the problem here?
Thanks
 A: Your approach did not work because the number of distinguishable arrangements depends on whether or not the selection of three letters contains a repeated letter.
We break the problem into cases, depending on whether or not there is a repeated letter.  
The word BHARAT has six letters, including two A's, one B, one H, one R, and one T.  
Three different letters:  The word BHARAT has five different letters.  We choose three of them, then arrange the chosen letters in order, which can be done in 
$$\binom{5}{3} \cdot 3! = P(5, 3)$$
ways.
Two different letters:  This can only occur if we use both A's.  Choose two of the three positions for the A's.  This leaves us with four choices for the remaining letter, which can only be placed in the remaining position in one way.
$$\binom{3}{2}\binom{4}{1}$$
Total:  $$\binom{5}{3} \cdot 3! + \binom{3}{2}\binom{4}{1}$$ 
A: Hint:
You seem to pick out $3$ out of the $6$ letters followed by repairing the double count of the indistinguishable $2$ letters A. 
Observe however that not all of the arrangements have $2$ letters A.
