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

asked in Numerical Ability by Boss (11.7k points)
edited by | 93 views

2 Answers

+2 votes
Best answer
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 $
answered by Boss (35.4k points)
selected by
0
In this CATI and TACI will be counted as 1 right ?

Since it is combination not permutation.
+1

 yes!

0

@Mk Utkarsh

Here asking for combination

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

+1

 yes combinations not permutations

0 votes
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
answered by Active (1.8k points)

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