How many 4 letter words can we make from BANANAS

Question

As the title suggests, I want to know the value of possible permutations without repetition given the word BANANAS

BA₁N₁A₂N₂A₃S

so for example to word A₁N₁A₂S is the A₃N₁A₂S and we want to exclude it from our total value.

How would I calculate this, I've tried 7*6*5*4 /(3!)(2!) but I think this value is too small.

Answer 1

Using exponential generating functions, since there are 3 A's, 2 N's, 1 B, and 1 S, we get

$g_e(x)=\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)\left(1+x+\frac{x^2}{2!}\right)(1+x)^2=1+4x+7x^2+\frac{43}{6}x^3+\frac{19}{4}x^4+\cdots$,

so there are $(4!)(\frac{19}{4})=\color{red}{114}$ words of length 4.

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Alternatively, we can consider the number of A's used:

1) 3 A's: There are 3 choices for the remaining letter, giving $4\cdot3=12$ cases

2) 2 A's: With 2 N's, there are 6 cases, and with BN, SN, or SB we get $3\cdot12=36$ additional cases

3) 1 A: with 2 N's, we get $2\cdot12=24$ cases, and with NBS we get $4!=24$ additional cases

4) 0 A's: this gives $4\cdot3=12$ cases

Therefore the total is given by $12+42+48+12=\color{red}{114}$ cases.

Answer 2

You can make a distinct four letter string from $\sf AAABNNS$ by selecting then arranging: $$\sf (AAA[B|S|N]) \mid (AANN) \mid ([AA|NN]BS) \mid ([AAN|NNA][B|S]) \mid (ANBS)$$

Count the ways for each case, then add.

• Triple and singleton (3 choice of singletons)
  • $\binom 11\binom 3 1\frac{4!}{3!~1!}$
• Two pair
• Pair and two singletons
• Four singletons

Answer 3

Okay, bring out the big and boring guns.

We can have:

No multiple letters. Thus words using B,A,N,S. There are $4!$ such words.

One pair of double letters. That letter can be be A or N, and for the remaining two letters we can omit (A or N), B, or A. So there are $2*3$ choices of letters and for each choice of letter there are $4*3$ ways to place the two non repeating letters. So $6*4*3$ such words.

Two pairs of A and N. There are ${4 \choose 2}$ choices to arrange AANN.

Three letters. A and then N,B,S for the remaining letter. That is $3$ chooses of letters and $4$ choices of where to put the one non repeating letter. $3*4$ total.

So there are $4! + 6*4*3 + {4\choose2} + 3*4 = 24 + 72 + 6 + 12 = 114$ total.