9

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.

Haani Al-Chalabi
  • 105

3 Answers3

13

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.

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.

user84413
  • 27,747
5

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
Graham Kemp
  • 133,231
5

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.

fleablood
  • 130,341
  • You correctly found that there are $\binom{4}{2}$ arrangements with two pairs, but $\binom{4}{2} = 6$. Also, there are only two N's, so the only way to use three identical letters is to use the three A's. You can choose their positions in $\binom{4}{3}$ ways and have three choices for filling the open position. – N. F. Taussig Sep 17 '16 at 00:39
  • Gore dang it. The first was a typo and second an oversight. I'll fix. – fleablood Sep 17 '16 at 02:08