How many 5 letter words can be formed from the word management if two alike letters are always together How many 5 letter words can be formed from the word MANAGEMENT if two alike letters are always together?
My approach was like this:
The letters M, N, A , E appear twice and G, T appear once.
So, the first case was where all the letters were different, thus, the number of words formed were:
6C5 . 5!(further permutation) = 720
the second case was where there were two groups of two alike letters and one different letter.
Thus, the number of words formed were:
2C1 . 4C2 . 3! = 72
the third case was where there were one group of two alike letters and three different letters.
Thus, the number of words formed were: 
4C1 . 5C3 . 4!= 960
thus, the total number of words were:
720 + 72 + 960 = 1752
but this answer is wrong as the Real answer is 1824.
Where was I wrong? Please explain.
 A: There are 4 double letters: M,N,A,E and 2 single letters G,T. 
To form 5 letter words in which two letters alike are together, there are possibilities as follows:


*

*the 5-letter word contains only single letters: $\binom{6}{5} \times 5!$

*the 5-letter word contains 1 double letters and 3 single letters: $\binom{4}{1} \times \binom{5}{3} \times 4!$

*the 5-letter word contains 2 double letters and 1 single letter: $\binom{4}{2} \times \binom{4}{1} \times 3!$
Total give you 1824.
A: Your only mistake was the case in which there are two double letters and a single letter.  As you observed, the four letters $M$, $N$, $A$, $E$ each appear twice.  We must select two of them to be the double letters, which we can do in $\binom{4}{2}$ ways.  Since there are six distinct letters in the word $MANAGEMENT$, whichever two we choose to use as the double letters, there four distinct letters left from which to choose the remaining letter, which we can do in $\binom{4}{1}$ ways.  We now have three objects to permute, the two double letters and the other letter we selected, which we can do in $3!$ ways.  Hence, in this case, there are 
$$\binom{4}{2}\binom{4}{1}3! = 144$$
possible arrangements.  
Notice that $$6! + \binom{4}{1}\binom{5}{3}4! + \binom{4}{2}\binom{4}{1}3! = 720 + 960 + 144 = 1824$$  
A: The miscalculation is here:

the second case was where there were two groups of two alike letters and one different letter. Thus, the number of words formed were: 
  2C1 . 4C2 . 3! = 72

I assume your ${2\choose 1}$ implies you are choosing from $G$ and $T$ and ${4\choose 2}$ implies you are choosing 2 letters from $M,N,A,E$. Your ${4\choose 2}$ is correct, since you will get two groups of two alike letters, consequently total $4$ letters. However, your ${2\choose 1}$ is not correct (it must be ${4\choose 1}$), because you can choose not only from $G$ and $T$, but also from the remaining two letters of $M,N,A,E$. Because, a single letter will be a different (unlike) letter. 
All the rest is fine.
