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

edited | 135 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 = $\Large \binom{3}{2}$ $= 3$

$\text{Total} = 5 + 18 +8 + 3 = 34$
by Boss (34.7k points)
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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 $^{5}C_{4}\times 4!$

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yes combinations not permutations

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@Satbir  What if it was permutations? Then CATI & TACI would not be counted as the same ? What difference does it make ? Please explain?

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That is the difference between Permutation and combination.

In permutation order matters and in combination order doesn't matter.

CATI and TACI will not be same in case of permutation.

Suppose you have 1 apple and 1 mango and we have to select two fruits then there is only one way i.e. pick mango and apple (or apple and mango.) (this is combination). Arrangement doesn't matter since our goal is to just pick the two fruits.

But if we have to  1 apple and 1 mango ,then we can arrange them in 2 ways.

first apple then mango.

first mango then apple.

(this permutation). Arrangement matters.

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
by Active (1.8k points)

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