How many 5 letter words can be formed from the word management if two alike letters are always together

Question

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.

Answer 1

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.

Answer 2

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 $$6! + \binom{4}{1}\binom{5}{3}4! + \binom{4}{2}\binom{4}{1}3! = 720 + 960 + 144 = 1824$$

Answer 3

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.