Aptitude - Permutation and Combination - Discussion

Now, let me explain in brief for those who don't find it easy.

Given word is 'CORPORATION'
Total Alphabets = 11.
vowels = 5 i:e(O,O,A,I,O)
Consonants = 11-5=6 i:e(C,R,P,R,T,N)

Now,treating vowels as 1 alphabet as asked in problm we hv=6+1=7
Also alphabet R comes twice .

Thus from "IMPORTANT FORMULAS" NO.4
For (C,R,P,T,N+(O O A I O))
p1=C(1) p2=R(2) p3=P(1) p4=T(1) p5=N(1) p6=(O O A I O)(1)

(p1+p2+p3+p4+p5+p6)= n i:e 1+2+1+1+1+1=7.

Thus,
n!/(p1!).(p2)!.....(pr!)= 7!/1!.2!.1!.1!.1!.1!=7!/2!=2520.

Similarly,
vowels (O,O,A,I,O) can be arranged in
p1=O(3) p2=A(1)p3=I(1)

p1+p2+p3 i:e= 3+1+1=5

5!/3!.1!.1!=5!/3!= 20.

Now the word 'CORPORATION' can be arranged in= 2520*20=50400 ways.

For @Baz and @Emmanuel .Its not 5/3 its 5!/3! (factorial ways).

The formula for the division of this is give as a result in Important Formulas prescribed in this site (Formula 4). But rather than the formula it is more of Division Logic. This Division logic comes more often in chapters of Prob and statistics.

" Logic - for the vowel letters OOAIO, consider only OOO. In 3 spaces only 1 combination can be made, so in 5 spaces (OOO_ _ )how many combination can be made? hence 5!/3!.

Division is cummulative subtraction ( therefore in 3! ways only 1 way is the right answer and other all 3! - 1 are wrong answers)

@Aarthi,

The combination is selection only whereas permutation is selection + arrangement.

For example:

A committee should be formed with 2 persons from A, B, C, therefore, the should be {AB, BC, CA} only not {AB, BC, CA, BA, BC, CA} because AB and BA both are the same combination.

Next, If the question is like committee should be formed with one president and one vice president the answer is {AB, BC, CA, BA, BC, CA} because, in the combination AB, A can be president and B can be vice president and vice versa. So both are different.

Why divide 7! by 2! ? because you can treat the problem as a permutation with subgroups of identical items. the general formula is nPn1,n2,n3... equals n! divided by n1!n2!n3!...

In this problem you have n = 7 letters (6 plus the vowel group). two letters are the same so n1 = 2. the rest are unique so the 5 other subgroups = 1. so you 7! divided by
2!1!1!1!1!1! . the answer as given simply didn't write out the 1! terms.

You can do either of the way either consonants first or the vowels first.

Just separate the Word CORPORATION along with the vowels i.e

CRPRTN: These are 6 consonants, as are is repeated two times you need to divide it by 2!

Hence, 6+1 = 7!/2! = 5040/2 = 2520.

1 was from the set of vowels.

(OOAIO) - These are 5 vowels which is represented as 1 set.

Now, from the Vowels we get.

5!/3! = 120/6 = 20.

i.e 2520*20 = 50400.

Every repeated vowels and consonants, you must divide them with the corresponding numbers if it. Like the one up there, O is repeated thrice (3 times) so you must divide them with 3!. Same goes to R which is repeated twice. That's they must be divided with 2!. After that you can just multiply both of permutation, 7! and 5! which were already divided by their repeated letters.

Hi.

In the above {CRPRTN (OOAIO)} (OOAIO can be treated as one letter).

Out of those 7 letters can be arranged in 7! ways. But there are 2 R's is there. They can exchange their places. That's why we are doing 7!/2!.

7! = 7*6*5*4*3*2*1 = 5040/2! = 2520.

In the vowels similarly 3 O's can exchange their places.

5!/3! = 5*4*3*2*1/3*2*1 = 20 ways.

= 2520*20 = 50400.

Firstly there are 11 words(5 vowels and 6 non vowels).
Consider 5 vowels =1 and remaining words are 6.
Add them (6+1=7).
In those 7 words count the Repeating words and divide to 7 (likewise 7fact÷2fct) and,
In those 5 vowels count the repeating words to and divide to 5 (5fact÷3 fact).

Multiply both of them :

(7fact÷2fact) * (5fact÷3fact).
= 50400.

CORPORATION Vowels Come Together.

Consonants CRPRTN ( R repeat twice).
Vowels OOAIO ( O repeat thrice )

We take 7!/2! * 5!/3!.

Note: 7! means the number of consonants + the vowels taken as "one".
that means 6 + 1 ( CRPRTN + ( OOAIO)).
7!/2! means R repeated twice, so we take 2!.
5!/3! means O repeated thrice, so we take 3! too.

Please one should help me out with this complicating question:

(a) How many four and five digits numbers can be formed using 2, 3, 4, 5, 6?
(b) How many will be greater than 5,000?
(c) How many will be odd numbers?
(d) How many will be even numbers?

PLEASE show workings thank you.