How many ways are there to arrange the 7 letters AAABBBB?

Question

A. 2^7
B. P(7, 7)
C. C(7, 7)
D. P(7, 3)/P(3, 3)
E. C(7, 3)
F. C(7, 4)
G. P(7, 3)/3!

E and F I'm pretty sure are one of the choices are there others? I'm thinking about D and/or G but not sure

Answer 1

A) $2^7 = 2*2*2*2*2*2*2$

B) $P(7,7) = \frac {7!}{7!} = 1$.

C) $C(7,7) ={7\choose 7} = \frac {7!}{7!(7-7)!} = \frac {7!}{7!*0!}=\frac {7!}{7!} = 1$.

D) $\frac {P(7,3)}{P(3,3)} =\frac {\frac {7!}{3!}}{\frac {3!}{3!}}=\frac {4*5*6*7}{1} = 4*5*6*7$

E) $C(7,3)={7\choose 3} = \frac {7!}{3!(7-3)!}=\frac {7!}{3!4!}= \frac {4*5*6*7}{1*2*3*4}= \frac {5*6*7}{1*2*3} = 5*7$.

F) $C(7,4) = {7\choose 4} = \frac {7!}{4!(7-4)!} = \frac {7!}{4!3!} = \frac {5*6*7}{1*2*3} = 5*7$.

G) $\frac {P(7,3)}{3!}= \frac {\frac {7!}{3!}}{3!}= \frac {4*5*6*7}{1*2*3} = 4*5*7$.

.....

So which of those are equal to each other and which is the answer?

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Consider it in two steps.

How many ways are there to arrange $7$ letters?

Answer: $7!$.

Now suppose you have an arrangement of $7$ letters so that positions: $j,k,l$ are occupied by the three different $A$s. Keeping all the non-As in their positions and requiring the $A$s to be in the positions $j,k,l$ but no caring which of the $3$ $A$s are in which position how many ways are there to rearrange the $A$.

Answer: $3!$.

Okay, so you know how many ways to arrange $7$ letters and for each of those, you know how many would be the same positions of the $A$s but with different $A$, how would you figure how many ways to arrange the $7$ letters if you consider the $A$s to be more or less the same letter? What would you do?

Answer: There $7!$ ways total but for every set of three positions for the $3$ $A$s there are $3!$ such arrangements of $A$s. Then there are $\frac {7!}{3!}$ different ways to choose the three positions for the $A$ (regardless of which $A$s go in the positions).

Okay so do the same for the $B$s in position.

There are $7!$ ways to arrange thee letters. For any positions of the $A$ and $B$s there are $3!$ ways to arrange the $A$ and $4!$ ways to arrange the $B$s and so there total numbe of ways to choose positions for the $A$s and $B$s, is $\frac {7!}{3!4!}$.

Answer 2

Hint:

Your problem is the same as having $7$ boxes, and you have $3$ A's to place. Maximum $1$ A per box. The rest of the remaining boxes get a B each.

How manys ways are there to place $3$ A's given $7$ boxes?

Answer 3

Meditate on The Tao of BOOKKEEPER, enlightening will come.

To summarize: Consider $A_1 A_2 A_3 B_1 B_2 B_3 B_4$, those are 7 different letters that can be permuted in $7!$ ways. Take any of those permutations, "turn off" the subindices of the $A$s, that means that $3!$ permutations are indistinguishable now, do the same with the $B$s, a factor $4!$ more. In all:

$$\frac{7!}{3! 4!} = \binom{7}{3}$$

This can here be explained as picking 3 positions out of 7 for the $A$s (or 4 out of 7 for the $B$s). In the case of BOOKKEEPER (1 $B$, 3 $E$, 2 $K$, 2 $O$, 1 $P$, 1 $R$) the result turns out to be:

$$\frac{10!}{1! 3! 2! 2! 1! 1!} = \binom{10}{1; 3; 2; 2; 1; 1}$$

a multinomial coefficient.