The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+1 vote
135 views

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

in Numerical Ability by Boss (10.8k points)
edited by | 135 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 $
by Boss (34.7k 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

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

+1

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.

 

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
by Active (1.8k points)

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
49,807 questions
54,727 answers
189,302 comments
79,844 users