Quantitative Aptitude
PERMUTATION AND COMBINATION MCQs
Permutations And Combinations
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
So, Required number of ways = = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= \(\left(\frac{7\times6\times5}{3\times2\times1}\times\frac{6\times5}{2\times1}\right)\) + (7C3 x 6C1) + (7C2)
= \(525+\left(\frac{7\times6\times5}{3\times2\times1}\times6\right)+\left(\frac{7\times6}{2\times1}\right)\)
= (525 + 210 + 21)
= 756.
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
So,, Required number of ways = (120 x 6) = 720.
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = \(\frac{7!}{2!}=2520.\)
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in \(\frac{5!}{3!}=20ways.\)
So, Required number of ways = (2520 x 20) = 50400.
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (7C3 x 4C2)
= \(\left(\frac{7\times6\times5}{4\times3\times1}\times\frac{4\times3}{2\times1}\right)\)
= 210.
Number of groups, each having 3 consonants and 2 vowels = 2
Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
So,Required number of ways = (210 x 120) = 25200.