How do you find the number of distinct arrangements of the letters in BOOKKEEPER?

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How do you find the number of distinct arrangements of the letters in BOOKKEEPER?

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Answer 1 · 2016-07-03T07:35:59

$\frac{10!}{2! 2! 3!} = 151200$ $(10!)/(2!2!3!) = 151200$

Explanation:

There are a total of $10$ $10$ letters.

If they were all distinguishable then the number of distinct arrangements would be $10!$ $10!$ . We can make them distinguishable by adding subscripts:

$B O_{1} O_{2} K_{1} K_{2} E_{1} E_{2} P E_{3} R$ $BO_1O_2K_1K_2E_1E_2PE_3R$

If we remove the subscripts from the letter $O$ $O$ 's, then it no longer makes any difference what order the $O$ $O$ 's are in and we find that $\frac{1}{2!} = \frac{1}{2}$ $1/(2!) = 1/2$ of our $10!$ $10!$ arrangements are identical to the other half.

So there are $\frac{10!}{2!}$ $(10!)/(2!)$ possible arrangements of the letters:

$B O O K_{1} K_{2} E_{1} E_{2} P E_{3} R$ $BOOK_1K_2E_1E_2PE_3R$

If we remove the subscripts from the letter $K$ $K$ 's a similar thing happens and we are left with half again. So there are $\frac{10!}{2! 2!}$ $(10!)/(2!2!)$ possible arrangements of the letters:

$B O O K K E_{1} E_{2} P E_{3} R$ $BOOKKE_1E_2PE_3R$

Finally, since $E_{1}$ $E_1$ , $E_{2}$ $E_2$ and $E_{3}$ $E_3$ can be arranged in $3!$ $3!$ possible orders, then when we remove the subscripts from the $E$ $E$ 's there are $\frac{10!}{2! 2! 3!}$ $(10!)/(2!2!3!)$ distinct arrangements of the letters:

$B O O K K E E P E R$ $BOOKKE EPER$

$\frac{10!}{2! 2! 3!} = \frac{10!}{2 \cdot 2 \cdot 6} = \frac{10!}{4!} = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 = 151200$ $(10!)/(2!2!3!) = (10!)/(2*2*6) = (10!)/(4!) = 10 * 9 * 8 * 7 * 6 * 5 = 151200$

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How do you find the number of distinct arrangements of the letters in BOOKKEEPER?

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