Test: Permutation, Combination & Probability - 2 (2017-2010) - UPSC MCQ

Considering the worst case First 15 picked balls are red and white. Hence in 16th time it is assured that 3 different colour balls exist.

Let number of a pairs of  brown socks = y
Price of brown socks = x
Price of black socks = 3x
According to question
⇒ 5 x 3x + yx = 100   ... (i)
Now, clerk has interchanged socks pairs then price is increased by 100%

⇒ 3xy + 5x = 30x + 2xy ... (ii)
⇒ 30x + 2yx = 3xy + 5x
⇒ 25x = xy
y = 25
∴ So, number of  brown socks = 25

It mathematics I is not opted, then two subjects out of four subjects have to be opted for.
∴ Number of ways in which two subjects can be opted for 
If mathematics II is opted, then it can be offered only if mathematics I is also opted for Number of ways in which two subjects can be opted for = 6 +1 = 7.

Number of ways to select Principal = ²C₁
Number of ways to select Vice Principal = ⁵C₂
Total number of ways = ²C₁ + ⁵C₂

Number of possible combinations of selectres = 2 × 10 = 20

Now, total number of Marbles = 5 + 10 + 13 = 28

Let x be the number of women.
Let y be the number of men.
Total number of hand shakes = xy = 24
Then, the possible factors of x and y are x = 6 or  4, y = 4 or 6
Number of hugs = ^xC₂+ ^yC₂

Here are five persons, and 5 tasks
So, When T₂ task is fixed for person 3

For Task 1 no. of ways = 2
Task 2 no. of ways = 1
Task 3 no. of ways = 3
Task 4 no. of ways = 3
Task 5 no. of ways = 3
Total no. of ways for condition = 3 + 3 + 3 + 2 + 1 = 12
Condition II
When task T₂ is given to be person 4

No. of ways for Task T₁ = 2
No. of ways for Task T₂ = 1
No. of ways for Task T₃ = 3
No. of ways for Task T₄ = 3
No. of ways for Task T₅ = 3
Total number of ways for condition II
= 3 + 3 + 3 + 2 +1 = 12
Total number of ways for condition I and II = 12 + 12 = 24

We cannot predict the number of dinners for a particular member from the given data. It may be possible that by choosing members from picking lots, one may have to host a dinner more than one times.

Let no. of column = x, no. of rows = y
∴ xy = 630 – [3 × 1 + 3 × 2 + ......... + 3 × (y – 1)]
= 630 – 3 [1 + 2 + ........ + (y – 1)]

(a) If y = 3, then 3x = 630 – 9 

(b) If y = 4, then 4x = 630 – 18 

(c) If y = 5, then 5x = 630 – 30 

Given, n (U) = 100
Number of students who play cricket = 60
i.e. n(C) = 60
Number of students who play football = 30
i.e. n(F) = 30
Number of students who play both the games = 10
i.e. n(C ∩ F) = 10
To find : n(C' ∩ F') = ?
we know,
n(C ∪ F) = n(C) + n(F) – n(C ∩ F)
= 60 + 30 –10 = 80
n(C' ∩ F') = n(C ∪ F)' = n(U) – n(C ∩ F)
= 100 – 80 = 20

Take, A and B to be always together as a single entity.
Now, total no. of children = 4 – 2 + 1 = 3
These can be arranged in 3! ways and A, B can be arranged among themselves in 2! ways.
Hence, no. of arrangements such that A and B are always together = 3! × 2! = 3 × 2 × 2 = 12

First person can shake hand with the other 9 i.e, in 9 ways. Second person can shake hand with the remaining 8 persons and so on.
∴ total no. of hands shaked = 9 + 8 + ....... + 2 + 1

Different sums of money can be formed by taking one, two, three and all the four notes together.
No. of different sums = ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄
= 4 + 6 + 4 + 1 = 15

Each question can be answered in 2 ways.
Hence, total no. of sequences = 2 × 2 × ........ 10 times.
= 2¹⁰ = 1024