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

1 Answer
Jul 3, 2016

Answer:

#(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:

#BO_1O_2K_1K_2E_1E_2PE_3R#

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:

#BOOK_1K_2E_1E_2PE_3R#

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:

#BOOKKE_1E_2PE_3R#

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:

#BOOKKE EPER#

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