How to find number of words made using letters of word 'EQUATION' if order of vowels do not change

Question

Find number of words made using letters of word 'EQUATION' if order of vowels do not change.

My attempt:- since we do not have to change the order of the vowels hence, _E_U_A_I_O_ we have $6$ places to fill the remaining letters. Therefore total number of cases $= C(6,3) \times 3! + 6 \times 3!= 156$ But the answer is $336$.

Answer 1

By using your approach with _E_U_A_I_O_, you have first to count the number of non-negative integer solutions of $$x_1+x_2+x_3+x_4+x_4+x_5+x_6=3$$ where $x_i$ is the $i$-th gap size. By Stars-and-bars we get $\binom{3+6-1}{6-1}=\binom{8}{5}$. Since the three consonants QTN to be placed in the gaps are different, this number has to be multiplied by $3!$ and we obtain $$\binom{8}{5}\cdot 3!=8\cdot 7\cdot 6=336.$$

I think it is easier if you change the counting strategy. The word "EQUATION" has $8$ letters. We choose first the places for the $5$ vowels EUAIO to arrange in order in $\binom{8}{5}$ ways, then we place the remaining three consonants QTN in $3!$ ways. Therefore we find again $336$.

Answer 2

You need to put Q, T and N into a blank eight-letter word, and fill in the remaining five places with the vowels in the prescribed order. There are $3!=6$ ways to order the consonants, and $\binom{6}{3}=56$ choices of three positions to put them in. So the answer is $6\times56=336$.

Answer 3

It's $$\binom{6}{3}3!+\binom{6}{1}\binom{5}{1}3!+\binom{6}{1}3!.$$

I think you missed the case, when two letters side by side and one separately.

Answer 4

There are simpler approaches.

Method 1: We place the consonants first.

Since the word EQUATION has eight letters, there are eight positions to fill. Therefore, there are eight ways to place the Q, seven ways to place the T, and six ways to place the N. Since the vowels must appear in the original order, there is only one way to place the five vowels in the remaining five positions. Hence, the number of arrangements of the letters of the word EQUATION in which the vowels appear in the original order is $8 \cdot 7 \cdot 6 = 336$.

Method 2: We use symmetry.

Since the word EQUATION has eight distinct letters, there are $8!$ ways to arrange its letters. Within a given arrangement, the five distinct vowels can be permuted among themselves in $5!$ ways. Of these $5!$ ways of arranging the vowels, only one leaves the vowels in the original order. Hence, the number of admissible arrangements of the letters is $$\frac{8!}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5!}{5!} = 8 \cdot 7 \cdot 6 = 336$$