MCQ: Permutations and Combinations - 2 - SSC CGL MCQ

ALLAHABAD = 9 letters. Out of these 9 letters there is 4 A's and 2 L's are there.
So, permutations 
(a) There are 4 vowels and all are alike i.e. 4A's.

These even places can be occupied by 4 vowels. In 
In other five places 5 other letter can be occupied of which two are alike i.e. 2L's.
Number of ways = 5!/2! Ways.
Hence, total number of ways in which vowels occupy the even places  = 60 ways.

(b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct.
These 8 letters can be arranged in = 8!/4! Ways.
Also two L can be arranged themselves in 2! ways.
So, Total no. of ways in which L are together = 1680 × 2 = 3360 ways.
Now,
Total arrangement in which L never occur together,
= Total arrangement - Total no. of ways in which L occur together.
= 7560 - 3360
= 4200 ways

Let the Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144

Number of ways

There 10 questions in part A out of which 8 question can be chosen as = ¹⁰C₈
Similarly, 5 questions can be chosen from 10 questions of Part B as = ¹⁰C₅
Hence, total number of ways,

APPLE = 5 letters.
But two letters PP is of same kind.
Thus, required permutations,

No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!

⁵C₁ × ³C₁ - 1
= 15 - 1
= 14

There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways is,

A triangle needs 3 points.
And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,

The number of triangle can be formed by 10 points = ¹⁰C₃
Similarly, the number of triangle can be formed by 4 points when no one is collinear = ⁴C₃
In the question, given 4 points are collinear, Thus, required number of triangle can be formed,
= ¹⁰C₃ - ⁴C₃
= 120 - 4
= 116

One digit positive numbers = 5
Two digit positive numbers = 25
Three digit positive numbers = 100
4 digit positive numbers = 300
5 digit positive numbers = 600
Six digit positive numbers = 600
Total positive numbers,
= 5 + 25 + 100 + 300 + 600 + 600
= 1630

There are 9 non-zero digits to arrange themselves at 4 different position. Each letter can be arrange at different position in 9 different ways.
So, required number of ways,
= 9 × 9 × 9 × 9
= 9⁴

Let the 3 boys be B₁, B₂, B₃ and 4 prizes be P₁, P₂, P₃ and P₄
Now B₁ is eligible to receive any of the 4 available prizes (so 4 ways)
B₂ will receive prize from rest 3 available prizes(so 3 ways)
B₃ will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways

Let n be the number of persons in the party
Number of hands shake = 105
Total number of hands shake is given by ⁿC₂
Now,
According to the question,

But, we cannot take negative value of n
So, n = 15
i.e. number of persons in the party = 15

The number of matches in first round,
= ⁶C₂ +⁶C₂
Number of matches in next round,
= ⁶C₂
Number of matches in semifinals,
= ⁴C₂
Total number of matches,
= ⁶C₂ + ⁶C₂ + ⁶C₂ + ⁴C₂ + 2
= 53