Number of words that can be formed from the letters of the word ENGINEER so that order of the vowels do not change?

Question

Number of words that can be formed from the letters of the word ENGINEER so that order of the vowels do not change?

My work : Since order of vowels does not change, the order should always be EIEE. So I assumed it to be a single object ( EIEE) and arranged it along with NGNR in $5!/2!$ ways. But in this case, the situation when first E is separated by rest IEE and many more like that are not included. So how do I involve all cases?

Answer 1

There are five positions to be filled with the letters $N,G,N,R$: $$ \text{_ E _ I _ E _ E _} $$ Step 1: Let $x_1,x_2,x_3,x_4,x_5$ be the number of letters in the relevant positions. Then: $$x_1+x_2+x_3+x_4+x_5=4, \\ 0\le x_1,x_2,x_3,x_4,x_5\le 4$$ Using stars and bars it is: $${4+5-1\choose 5-1}={8\choose 4}.$$ Step 2: Permutation of the four letters $N,G,N,R$ is: $$\frac{4!}{2!}.$$ Hence, the total number of words is: $${8\choose 4}\cdot \frac{4!}{2!}=840.$$

Answer 2

You should simply count the number of arrangements of the letters N-G-N-R among eight positions, i.e. $(8!/(4!4!))*(4!/2!)$.

Answer 3

Here are two additional methods:

Method 1: We use symmetry.

The word $ENGINEER$ has eight letters, of which three are $E$s, two are $N$s, one is a $G$, one is an $I$, and one is an $R$. We can choose three of the positions for the $E$s, two of the remaining five positions for the $N$s, and arrange the remaining three distinct letters in the three remaining spaces in $$\binom{8}{3}\binom{5}{2}3!$$ ways.

By symmetry, one fourth of these arrangements will have an $I$ in the second position among the four vowels. Hence, the number of admissible arrangements is $$\frac{1}{4}\binom{8}{3}\binom{5}{2}3!$$

Method 2: We place the consonants, then the vowels.

We choose two of the eight positions for the two $N$s, one of the remaining six positions for the $G$, and one of the remaining five positions for the $R$. There is only one way to arrange the vowels in the remaining four positions so that they appear in the same order that they do in the word $ENGINEER$. Hence, the number of admissible arrangements is $$\binom{8}{2}\binom{6}{1}\binom{5}{1}$$