InfyTQ Permutation and Combination Quiz 2

Following would possible scenarios: Men - (0, 1, 2, 3)

If there are 0 men, then 11 women. Similarly, if there are 1 men then there would be 10 women so on ...

• Ways of choosing (0 Men, 11 Women) : (⁵C₀ × ¹¹C₁₁)


• Ways of choosing (1 Men, 10 Women) : (⁵C₁ × ¹¹C₁₀)


• Ways of choosing (2 Men, 9 Women) : (⁵C₂ × ¹¹C₉)


• Ways of choosing (3 Men, 8 Women) : (⁵C₃ × ¹¹C₈)

with replacement =\frac{8}{13}\times \frac{8}{13}\times {8}{13}=\frac{512}{2197}
without replacement = \frac{8}{13}\times \frac{7}{12}\times \frac{6}{11}=\frac{336}{1716}

First we have to select a bag and then we will draw a ball.
Probability of selection of both bags is equal =\frac{1}{2}
Now probability of blue ball taken from first bag = ( \frac{1}{2}) \times (\frac{3}{11})
and probability of blue ball taken from second bag =(\frac{1}{2}) \times (\frac{4}{11})
So probability of blue ball = (\frac{1}{2}) \times \frac{3}{11}+ (\frac{1}{2}) \times \frac{4}{11} =\frac{7}{22}

3m+4a=85...(i)
5a+6p=122....(ii)
6m+2p=114....(iii)
Adding (i) (ii) (iii) we get
9a+9m+8p= 321
9a+9m+9p= 321+p
a+m+p=(321+p)/9------(iv)

It must have to be a integer number.
To make (iv) as an integer p must have to be either 3 or (3+9) or (3+18)or..... as 324 or 333 or 342 is divisible by 9
Let check it out..
if p=3 then
5a+18=122
5a=104
a!= an integer so p=3 is false
If a=12, then
by(ii)
5a=122-72
a=10
by(i)
3m=85=40
a=15
a+m+p= 15+10+12=37.

or

A----- => 5!=120
AD---- => 4!=24
AED--- => 3!=6
AEI--- => 3!=6
AEO--- => 3!=6
24+6+6+6=42
AERDIO => 43th
AERDOI => 44th

we have 4 couple
1)first two husbands goes to another side
2)then one husband comes back
3)then one couple will go
4)then second husband comes back
so in this 4 ways 1 couple reach to another side
so for 4 couples there are 16 ways
a)16 is anss

He had 76 rupees,according to questions, he remember any no. larger
than 15 as remainder when divided by 3,5 and 7 respectively.
Divide 76 by 3,5 and 7, we will get (1,1,6)as remainder.
He paid 20 rupees bcoz when 20 divided by 3,5 & 7 , we will get (2,0,6) as
remainder.
Now money left=76-20
=56 (Ans)

let total no of balls =x
blue=\frac{2x}{3}
pink=\frac{x}{3}
total no of defective balls = \frac{10x}{27} + \frac{7x}{24}
=\frac{143x}{216}
non defective balls=x-\frac{143x}{216}=146 ,

x=432

No of ways that 2 chips selected from total of 26 is

^{26}\textrm{C}_{2}

=26\times \frac{25}{2} = 13\times 25
No of ways that 1 white and 1 blue is selected is ^{3}\textrm{C}_{1}\times^{7}\textrm{C}_{1} = 21
total probability = \frac{21}{13 \times 25}

the least possible no of member of members on any one committee = 4 .. option a)

In all 3 committees, say X,Y,Z, 2 persons say A and B are common.
C is common between X and Y.
D is common between Y and Z.
E is common between X and Z.

so X committee has ABCE. Y committee has ABCD. Z committee has ABDE.