Permutations and Combinations: Question 3

Worked solution

Part (a): Total arrangements of the 10 letters

The word BOOKKEEPER has $10$ letters: B, O, O, K, K, E, E, E, P, R. Reading off the letter counts: O appears $2$ times, K appears $2$ times, E appears $3$ times, and B, P, R each appear once.

If all $10$ letters were different, there would be $10!$ arrangements. Since swapping identical letters with each other doesn’t create a new arrangement, divide by the factorial of each repeated count:

$\frac{10!}{2!\,2!\,3!} = \frac{3628800}{2\times2\times6} = \frac{3628800}{24} = 151200.$

Check (positions method): choose which $3$ of the $10$ positions hold an E: $\binom{10}{3}=120$. From the remaining $7$ positions, choose which $2$ hold an O: $\binom{7}{2}=21$. From the remaining $5$ positions, choose which $2$ hold a K: $\binom{5}{2}=10$. The last $3$ positions take the distinct letters B, P, R in any of $3!=6$ orders. $120\times21\times10\times6 = 151200.$ Both methods agree: $151200$.

Part (b): The two Ks together (block method)

Glue the two Ks into a single block, KK. Because the two Ks are identical, there is no extra factor for “ordering inside the block”. Swapping them gives the same block.

This leaves $9$ objects to arrange: the block KK, together with B, O, O, E, E, E, P, R (that’s $8$ single letters $+$ $1$ block $=9$ objects). Among these, O still repeats twice and E still repeats three times:

$\frac{9!}{2!\,3!} = \frac{362880}{2\times6} = \frac{362880}{12} = 30240.$

Check (proportion argument): among all $\binom{10}{2}=45$ ways to choose the $2$ (identical) positions for the two Ks out of $10$ slots, the number of ways that give adjacent positions is $9$ (the pairs $(1,2),(2,3),\ldots,(9,10)$). So the fraction of arrangements with the Ks adjacent should be $\frac{9}{45}=\frac{1}{5}$. Checking against part (a)‘s total: $151200\times\frac{1}{5} = 30240.$ Both methods agree: $30240$.

Part (c): The two Ks not together (complement)

The arrangements with the Ks not adjacent are the total from part (a) minus the “together” arrangements from part (b):

$151200 - 30240 = 120960.$

Check (gap method): first arrange the $8$ non-K letters (B, O, O, E, E, E, P, R), which has $\dfrac{8!}{2!\,3!}=\dfrac{40320}{12}=3360$ arrangements. This creates $9$ gaps (one before, one after, and one between each pair of these $8$ letters) into which the $2$ identical Ks can be placed so they are never adjacent to each other. Choosing $2$ of these $9$ gaps: $\binom{9}{2}=36$. $3360\times36 = 120960.$ Both methods agree: $120960$.

Final answers

• (a) Total arrangements $= \boxed{151200}$
• (b) The two Ks together $= \boxed{30240}$
• (c) The two Ks not together $= \boxed{120960}$