Tech Mahindra Permutation & Combination Quiz 1

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (7C3 x 4C2)
= 7 x 6 x 5 x 4 x 3
3 x 2 x 1 2 x 1
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
Required number of ways = (210 x 120) = 25200.

physics book can be arranged in 5!=120
both the book can be arranged in 2!=2 eg (MMM)(PPPPP) or (PPPPP)(MMM)
so total arrangement will be 2!*5!*3! = 2*120*6 = 1440

(d) Number of ways of selecting 3 points from given 12 points = ¹²C₃
But any three points selected from given seven collinear points does not form triangle.
Number of ways of selecting three points from seven collinear points = ⁷C₃ Required number of triangles = ¹²C₃ – ⁷C₃ = 220 -35 = 185 .

If there is one letter and one envelope then no way you can put it wrong(S1).
If there are 2 letters and 2 envelopes then you can put them wrong in 1 way(S2).
If there are 3 letters and 3 envelopes then you can put them wrong in 2 ways(S3).
If there are 4 then you can put them wrong in 9 ways(S4).
If there are 5 then you can put them wrong in 44 ways(S5).
If you observe you can find a pattern.
S3=(S1+S2)*2
S4=(S2+S3)*3
S5=(S3+S4)*4
S6=(S4+S5)*5
In general, Sn=(Sn-2 + Sn-1)*(n-1)
So, if there are 5 letters then S5=(S3+S4)*4=(2+9)*4=44

4 flowers which are always together can be considered as one SET,
Therefore we have to arrange one SET ( 4 flowers ) and 4 other flowers into a garland.
Which means, 5 things to be arranged in a garland.
(5-1)!
And the SET of flowers can arrange themselves within each other in 4! ways.
Therefore
(5-1)!*(4!)
But, Garland, looked from front or behind does not matter. Therefore the clockwise and anti clockwise observation does not make difference.
Therefore
(5-1)! * (4!)/ 2
= 288.

A voter can either give One or Two or Three votes
No. of ways to give one vote : 5C1=5
No. of ways to give two votes : 5C2=10
No. of ways to give three votes : 5C3=10
No. of ways to give three votes : 5C3=10
A candidate may not vote to anyone=1
Hence, Total ways = 5+10+10-1 = 25-1 = 24 ways

Expert
We have ‘p’ copies of each of ‘n’ different books.
Number of ways to select from p copies:
p + 1
Total no. of ways:
(p + 1)(p + 1))p + 1)................n times
(p + 1)n
Number of different ways of selection is
(p + 1)n– 1

(3C1*4C1*5C1)+(3C2*4C1)+(3C2*5C1)+(4C2*3C1)+(4C2*5C1)+(5C2*3C1)+(5C2*4C1)
=60+12+15+18+30+30+40
=205

Let the 9 friends be A, B, C, D, E, F, G, H and i respectively.
A and B do not attend the party together.
Total number of ways to select 5 from 7 = 7C5
= (7*6)/(2*1)
= 42/2
=  21 ways
Either of A or B is selected for party, then number of ways = 2C1*7C4
= (2*1)*(7*6*5)/(3*2*1)
= 420/6
= 70 ways
Total number of ways = 21 + 70
= 91 ways

10c6=210
captain can be selected from 6 player in 6 ways
210*6=1260