How many 4 letter words can we make from BANANAS 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.
 A: 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

A: 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.
A: 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.
