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.
Best Answer
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.