Applications of Permutations And Combinations - JEE MCQ

The total number of ways is same as the number of arrangements of 4 fingers, taken 3 at a time.
So, required number of ways = ⁴P₃ 
= 4!/(4-3)!
= 4!/1!
= 4! => 24

The person has 8 options to enter the hall. For each of these 8 options, he has 7 options to exit the hall. Thus, he has 8 × 7 = 56 ways to enter and exit from different doors.

There are five questions
Each question has 4 options
No. of possible ways of answering each question is four
No. of Possible ways for Q1 = 4
No. of Possible ways for Q2 = 4
No. of Possible ways for Q3 = 4
No. of Possible ways for Q4 = 4
No. of Possible ways for Q5 = 4
So, Total number of ways of answering 5 objective type questions, each question having 4 choices = 4⁵
= 1024

Lets first place the men (M). '*' here indicates the linker of round table
 
* M -M - M - M - M *
which is in (5-1)! ways
So we have to place the women in between the men which is on the 5 empty seats ( 4 -'s and 1 linker i.e * )
So 5 women can sit on 5 seats in (5)! ways or
1st seat in 5 ways
2nd seat 4
3rd seat 3
4th seat 2
5th seat 1
i.e 5*4*3*2*1 ways
So the answer is 5! * 4! = 2880

Since the exterior angle 140 degrees, The sum of the interior angles = (2n - 4)* right angles. So 140n = (2n - 4)* right angles, or
140n = (2n - 4)*90, or
140n = 180n - 360o, or
40n = 360°, or
n = 9 sides.

Total no of players = 14 out of which 5 are fixed.
So, 11-5 = 6
Remaining players = 14 - 6
= 9 players
9C6 = 9!/(3!*6!)
= 84 

No. of ways to select 4 senior and 3 U-19 players = ⁶C₄ * ⁵C3 = 150
No. of ways to select 5 senior and 2 U-19 players = ⁶C₅ * ⁵C2 = 60
No. of ways to select 6 senior and 1 U-19 players = ⁶C₆ * ⁵C1 = 5 
Total no. of ways to select the team = 150 + 60 + 5 = 215

Total number of letters = 4
 Number of ordered pairs of letters that can be formed like (A, M) or (P, O) etc = ⁴P₂ ​= 4!/2!
​= 24/2
​= 12

There are 8 letters in the word TRIANGLE.
2 alphabets are fixed, remaining are 6 alphabets
So, number of arrangements = ⁶P₆
= 6!
= 720

Total no of balls = 9
red balls = 4
yellow balls = 3
green balls = 2
Total no. of arrangements = 9!/(4!*3!*2!)
= 1260