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

Jul 3, 2016

(10!)/(2!2!3!) = 151200

#### Explanation:

There are a total of $10$ letters.

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

$B {O}_{1} {O}_{2} {K}_{1} {K}_{2} {E}_{1} {E}_{2} P {E}_{3} R$

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

So there are (10!)/(2!) possible arrangements of the letters:

$B O O {K}_{1} {K}_{2} {E}_{1} {E}_{2} P {E}_{3} R$

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

$B O O K K {E}_{1} {E}_{2} P {E}_{3} R$

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

$B O O K K E E P E R$

(10!)/(2!2!3!) = (10!)/(2*2*6) = (10!)/(4!) = 10 * 9 * 8 * 7 * 6 * 5 = 151200