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.

I have the same question, Is there a formula behind ?

I also can't understand division. Is there any formula ?

Here we can simply separate out the common alphabet.

in CORPORATION , o is 3 times and r is two times, putting together we have,OOORRCPATIN . now the total unique alphabets are 7 and thus the answer is 7%=7*6*5*4*3*2*1= 50400 . the common alphabets dun need any permutation.

Actually if you consider ooo as o and rr as are the there are 8 distinct alphabets. C O R P A T I N.

How the 7 is came instead of 6 (CRPRTN).

What is process when same letter is come in like as this question?

Thanks.

We consider all vowels as one letter i.e. OOAIO is a letter + rest of the 6 letters = 1+6 =7.

Good explanation. Thank you.

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.