Test: Permutation & Combination- 3 - CAT MCQ

The number is to be of 9 digits
The first place can be filled in 9 ways only (as 0 can not be in the left most position )
Having filled up the first place the remaining 8 places can be filled in 9×8×7×...×1=9! ways
Hence total number of 9 digit numbers with distinct digits is =9×9!

Given, x+y+z+u+t=20...........(1)

and x+y+z=5................(2)

The given system of equations can be written as

u+t=15........(3)

x+y+z=5.............(4)

No. of non-negative integral solutions of (4) are ^n+r−1​C_r​=³⁺⁵⁻¹​C_5​=^7​C₅​

No. of non-negative integral solutions of (3) are ^n+r−1​C_r​=²⁺¹⁵⁻¹​C₁₅​=¹⁶​C₁₅​

So, required numbers =^16​C₁₅​×⁷C₅​=336

ⁿC₃ - ^n-1C₂ = 84, where n is the number of students. Now use options.

H ere, Since x ≥ 0 , y ≥ l , z ≥ 2

Now, x can get any number of things, y should get a minimum of 1 thing and z should get a minimum of 2 things. So, let us give these to y and z.
Now , total things left = 12

Now use the formula n identical things can be distributed among r persons in ^n+x-1C_x-1 

Where n = 12, r = 3

Various ways of doing this is as follows:

Now do the selections.

This is nothing but ⁴C₂ x ³C₂.

This can be done in three ways:

i. A is selected but B is not selected - A has already been selected, rest 4 is to be selected from 7 persons = ⁷C₄ = 35

ii. B is selected but A is not selected - ⁷C₄ = 35

iii. Neither A nor B is selected - ⁷C₅ = 21 Total = 91

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 .

A chess board is in the shape of a square divided into 8*8 smaller squares of equal dimensions. 

If a rectangle formed has one side length 1, then following (length, breadth) combinations possible:- 

(1,1), (1, 2), (1, 3), (1, 4), 1,5), (1, 6), (1, 7), (1, 8) 

Total 8 non - congruent rectangles. 

Similarly with one side as 2 following non - congruent triangles possible:- 

(2, 2), (2, 3),.......(2, 8) . 
Total 7 non - congurent rectangles. 

Similarly for 3,4,5,6,7,8  we get 6,5,4,3,2,1  non - congurent rectangles respectively. 

So, total no. Of rectangles will be:- 

=> 8+7+6+5+4+3+2+1 

=> 36

. 1. As the first step, let calculate how many triangles can we have connecting all vertices of the decagon. The number of these triangles is exactly equal to the number of combinations of 10 items (vertices) taken 3 at a time  = 10*3*4 = 120.

2. Now let exclude from this amount those triangles that have ONLY ONE side of the decagon as their side. The number of excluded triangles is equal to 10*6 = 60 (10 sides of the decagon, and each side may go with one of (10-2-2) = 6 opposite vertices).

3. Last step is to exclude those triangles from 120 of the n.1, that have TWO SIDES of the decagon as their sides. The number of such triangles is 10 (exactly as the number of the decagon's vertices).

4. So, final answer is 120 - 60 - 10 = 50 triangles.

Number of the integral solution for abc=30 are:

1×3×10⇒Permutation=3!

15×2×1⇒Permutation=3!

5×3×2⇒Permutation=3!

5×6×1⇒Permutation=3!

30×1×1⇒Permutation= 3!/2!

Total solutions =(3!×4)+3=27

• Number of ways to choose at least one starter out of 5: 2^5 - 1 = 31 ways


• Number of ways to choose 6 main courses: 1 way since they are identical


• Number of ways to choose 4 desserts: 4! = 24 ways

• Total number of ways = Number of ways to choose starters x Number of ways to choose main courses x Number of ways to choose desserts


• Total number of ways = 31 x 1 x 24 = 31 x 24 = 744 ways

• Therefore, the correct answer is A: 31 x 6 x 15