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

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?

• So, you have to fill in the blanks given $_E_I_E_E_$. That's five blanks and you have four letters to use. So, first you need to count the five tuples of non-negative integers that sum to four. Now you have to permute $NGNR$ and divide accordingly. – lulu Sep 4 '18 at 13:25
• That will lead to 5C4 . 4!/2=60. Am I right? – Rohit Sinha Sep 4 '18 at 13:27
• There are more than $\binom 54$ five-tuples. That's only $5$. There are already five ways to do it if you put all the letters in one slot. – lulu Sep 4 '18 at 13:29
• I will be glad if you put some of the cases in the answer – Rohit Sinha Sep 4 '18 at 13:31
• In this case the numbers are so small that you can do it all out. In general, with larger numbers, a technique like Stars and Bars is useful. – lulu Sep 4 '18 at 13:32

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: $${8\choose 4}\cdot \frac{4!}{2!}=840.$$

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

• The method described by @lulu does work. First, you choose how many of the four letters go in each of the five spaces, then you arrange them in those spaces. Done properly, you obtain $\binom{8}{4}\binom{4}{2}2!$, which is equal to the answer you obtained. – N. F. Taussig Sep 4 '18 at 16:28
• Thanks for your remark. I'm afraid I read the comment too fast. – Mickybo Yakari Sep 4 '18 at 17:18

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 $$\frac{1}{4}\binom{8}{3}\binom{5}{2}3!$$
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 $$\binom{8}{2}\binom{6}{1}\binom{5}{1}$$