+1 vote
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How many 4 letter combinations can be made with the help of letters of the word STATISTICS?

edited | 93 views

S S S
T T T
A
I I
C

There can be 4 kinds of combinations
1. All different letters = $\Large \binom{5}{4}$ $= 5$

2. 2 same , 2 different = $\Large \binom{3}{1} . \binom{4}{2}$ $= 18$

3. 3 same, 1 different = $\Large \binom{2}{1}. \binom{4}{1}$ $= 8$

4. 2 same , 2 same = $\Large \binom{3}{2}$ $= 3$

$\text{Total} = 5 + 18 +8 + 3 = 34$
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0
In this CATI and TACI will be counted as 1 right ?

Since it is combination not permutation.
+1

yes!

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@Mk Utkarsh

So, 1st one will be $^{5}C_{4}\times 4!$

+1

yes combinations not permutations

This can be found out using generating function

$\text{A - 1, C - 1, I - 2, S - 3, T - 3}$

$\underbrace{A}\ \ + \ \underbrace{C} \ \ \ + \ \underbrace{I}\ \ \ \ + \ \underbrace{S}\ \ \ \ + \ \underbrace{T}\ = 4$

$\leq 1 \ \ \ \ \ \ \leq 1 \ \ \ \ \ \ \ \ \ \ \leq2 \ \ \ \ \ \ \ \ \ \ \leq 3 \ \ \ \ \ \ \ \ \ \ \leq3$

Now we need to find those coefficients whose power is 4

$\Rightarrow (1+x)(1+x)(1+x + x^2)(1+x+ x^2+x^3)(1+x+ x^2+x^3)$

$\Rightarrow (1+x)^2 \dfrac{(1-x^3)}{(1-x)} \dfrac{(1-x^4)}{(1-x)} \dfrac{(1-x^4)}{(1-x)}$

solving this equation we will get $34$ as answer

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