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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?

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    $\begingroup$ 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. $\endgroup$ – lulu Sep 4 '18 at 13:25
  • $\begingroup$ That will lead to 5C4 . 4!/2=60. Am I right? $\endgroup$ – Rohit Sinha Sep 4 '18 at 13:27
  • $\begingroup$ 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. $\endgroup$ – lulu Sep 4 '18 at 13:29
  • $\begingroup$ I will be glad if you put some of the cases in the answer $\endgroup$ – Rohit Sinha Sep 4 '18 at 13:31
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    $\begingroup$ 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. $\endgroup$ – lulu Sep 4 '18 at 13:32
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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: $${8\choose 4}\cdot \frac{4!}{2!}=840.$$

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

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  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Sep 4 '18 at 16:28
  • $\begingroup$ Thanks for your remark. I'm afraid I read the comment too fast. $\endgroup$ – Mickybo Yakari Sep 4 '18 at 17:18
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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 $$\binom{8}{2}\binom{6}{1}\binom{5}{1}$$

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