# How many ways are there to arrange the letters of word ALGEBRA such that only the relative order of the vowels does not change???

I have to arrange the word ALGEBRA but vowels must maintain their original relative order. So I thought we have 3 spaces if we put vowels like A_ _ E _ _ A (I know there are 3! ways to arrange those) but how do I approach the problem about consonants?

Could someone help me?

• I think "original relative order" means they need to be A-E-A but they can be placed anywhere as long as that is their order. Anyone else read it that way? – The Count Mar 8 '17 at 20:34
• Why $3!$? You have four consonants $L,G,B,R$ and four slots so surely it's $4!$. Of course this is only one pattern. You could have _ _ _ $AEA$, just to name one. – lulu Mar 8 '17 at 20:34
• Ok I suppose I didn't fully understand! Thanks :) – mandella Mar 9 '17 at 8:04

Imagine Alice, Elizabeth, and Anne standing in that order. Now send Larry to join them. He can fit in any of $4$ positions. Next send George, who can fit in any of $5$ positions, then Bruce, who'll have $6$ choices of where to fit, and finally Roscoe, who'll have $7$ choices. The total number of orderings is thus

$$4\cdot5\cdot6\cdot7=840$$

It doesn't really matter in what order you send the guys; the first guy will have $4$ choices, the next guy $5$, and so on.

Presumably preserving "the relative order of the vowels" means that their actual positions can change.

The vowel locations can be chosen in $\binom 73 = 35$ ways (see binomial coefficients).

The consonants can be arranged in $4!=24$ ways.

For total options, these are independent choices - the consonant ordering does not affect the vowel locations or vice versa - so multiply together: $35\cdot 24 = 840$

Method 1: We place the consonants first.

Since ALGEBRA has seven letters, we have seven positions to fill. If we place the B, G, L, and R in that order, we have seven choices where to place the B, six choices where to place the G, five choices where to place the L, and four choices where to place the R. Once the consonants have been placed, there is only one way to arrange the vowels in the remaining positions so that the E appears between the two A's. Hence, there are $$7 \cdot 6 \cdot 5 \cdot 4$$ ways of arranging the letters of the word ALGEBRA so that the relative positions of the vowels are preserved.

Method 2: We place the vowels first.

We choose three of the seven positions for the three vowels. There is only one way to place the vowels in these positions that preserves their relative order. This leaves four positions. We can arrange the four consonants in these positions in $4!$ ways. Hence, the number of ways of arranging the letters of the word ALGEBRA that preserves the relative order of the vowels is $$\binom{7}{3} \cdot 4!$$ Note that this is essentially a rephrasing of @Joffan's solution.

Method 3: We use symmetry.

First, we count the total number of distinguishable arrangements of the letters of the word ALGEBRA. We choose two of the seven positions for the A's. Once the A's have been placed, we can arrange the remaining five distinct letters in the remaining five positions in $5!$ ways. Hence, the number of distinguishable arrangements of the letters of the word ALGEBRA is $$\binom{7}{2} \cdot 5!$$ By symmetry, in one third of these arrangements does the E appear somewhere between the two A's. Hence, the number of arrangements of the word ALGEBRA in which the relative order of the vowels is preserved is $$\frac{1}{3} \binom{7}{2} \cdot 5! = 840$$

Another method is through distribution of the letters, then permuting it.

You know that it must look in this form; _A_E_A_, and you have to distribute the remaining 4 letters into any 4 "boxes". We first assume the letters are indistinguishable. By the distribution method, we have ${4+4-1 \choose 4}={7 \choose 4}$. After distributing the letters, we now start to permute the letters $(L, G, B, R) = 4!$.

Hence, the answer is ${7\choose4}.4!$, which is the same as the answers above. Personally, I like to use to distribution method, it really helps for many of the questions I've seen so far!