Number of words which are equivalent to BACHELOR

Question

Consider all the $8$-letter words that can be formed by arranging the letters in $BACHELOR$ in all possible ways. Any two such words are equivalent if those two words maintain the same relative order of the letters $A$, $E$ and $O$. For example, $BACOHELR$ and $CABLROEH$ are equivalent. How many words are there which are equivalent to $BACHELOR\;?$

I just try to find all possible cases with maintaining the order of $A$, $E$ and $O$, but their are lots of cases and I can't find the pattern. That's why I can't calculate all possible cases.

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Answer 1

There are $8!$ possible permutations if we ignore the restriction on the order. There are $6$ equivalence classes (as there are $6$ ways to order three letter) and each equivalence class contains the same number of permutations (there is an obvious bijection between any two equivalence classes). Thus the answer is $$\frac {8!}6=6720$$

Answer 2

Let $a_1$ denote the number of letters on left side of $A$.

Let $a_2$ denote the number of letters between $A$ and $E$.

Let $a_3$ denote the number of letters between $E$ and $O$.

Let $a_4$ denote the number of letters on right side of $O$.

Then the $a_i$ are nonnegative integers that satisfy $a_1+a_2+a_3+a_4=5$ and with inclusion/exclusion we find that there are $\binom{5+3}3$ of such sums.

There are $5!$ ways to arrange the letters $BCHLR$ so there are $$\binom835!$$ possibilities.

Answer 3

Method 1: There are $8$ ways to place the B, $7$ ways to place the C, $6$ ways to place the H, $5$ ways to place the L, and $4$ ways to place the R. The letters A, E, O must be placed in the open positions in that order. Hence, there are $$8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$$ such arrangements.

Method 2: Choose three of the eight positions for the vowels A, E, O. They can be placed in the chosen positions in one way. The remaining positions can be filled with the five consonants in $5!$ ways. Hence, the number of possible arrangements is $$\binom{8}{3}5!$$

Method 3: There are $3!$ ways to arrange the vowels A, E, O within a given arrangement of the letters of the word BACHELOR. Only one of these is permissible. There are $8!$ arrangements of the eight distinct letters of the word BACHELOR. Therefore, by symmetry, the number of permissible arrangements is $$\frac{1}{3!} \cdot 8!$$

Answer 4

In general for an anagram with repeated letters, you can use a multinomial coefficient . So this would be suitable for example for anagrams of "CARAVANS", where we have $8$ letters with one of them repeated $3$ times, so the multinomial coefficient is

$\dbinom{8}{3,1,1,1,1,1} = \dfrac{8!}{3!\cdot 1!\cdot 1!\cdot 1!\cdot 1!\cdot 1!} = \dfrac{8!}{3!} = \dfrac{40320}{6} = 6720$

This is exactly relevant to this case as you can ignore the difference between the vowels for the purpose of finding arrangements. Once an arrangement is found - say CvBLRvvH, with v being an anonymous vowel - there is only one way to make that equivalent to BACHELOR, allocating the vowels into the v spots in the order AEO to give CABLREOH.