Permutation And Combination Test

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Permutation And Combination Test - 3

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Weekly Quiz Competition

Question 1
3 / -1

The total number of 9-digit numbers of different digits is

A
10 (9!)

B
8 (9!)

C
9 (9!)

D
None of these

Solution

It will be a number of 9 digits. 9 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 = 9 x 9!
Question 2
3 / -1

Find the number of non-negative integral solutions to the system of equations x + y + z + u + t = 20 and x + y + z = 5.

A
220

B
336

C
44

D
528

Solution

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−1C_r=³⁺⁵⁻¹C₅=⁷C₅
No. of non-negative integral solutions of (3) are ^n+r−1C_r=²⁺¹⁵⁻¹C₁₅=¹⁶C₁₅
So, required numbers =¹⁶C₁₅×⁷C₅=336
Question 3
3 / -1

Ateacher takes 3 children from her class to the zoo at a time as often as she can, but she does not take the same three children to the zoo more than once. She finds that she goes to the zoo 84 times more than a particular child goes to the zoo. The number of children in her class is

A
12

B
10

C
60

D
None of these

Solution

ⁿC₃ - ^n-1C₂ = 84, where n is the number of students. Now use options.
Question 4
3 / -1

If x, y, and z are integers and x ≥ 0, y ≥ 1, z ≥ 2, x+y + z= 15 then the number of values of the ordered triplet (x, y, z) is

A
91

B
455

C
¹⁷C₁₅

D
None of these

Solution

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
Question 5
3 / -1

In an examination, the question paper contains three different sections A, B and C containing 4, 5 and 6 questions respectively. In how many ways, a candidate can make a selection of 7 questions, selecting at least two questions from each section?

A
890

B
1420

C
2700

D
3140

Solution

Various ways of doing this is as follows:
Now do the selections.
Question 6
3 / -1

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines, is

A
6

B
18

C
12

D
9

Solution

This is nothing but ⁴C₂ x ³C₂.
Question 7
3 / -1

A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two particular friends A and B will not attend the party together is

A
56

B
126

C
91

D
None of these

Solution

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
Question 8
3 / -1

If a denotes the number of permutations of x + 2 things taking all at a time, b the number of permutations of x things taking 11 at a time and c the number of permutations of x - 11 things taking all at a time such that a = 182be, then the value of x is

A
15

B
12

C
10

D
18

Solution
Question 9
3 / -1

How many 10 digits numbers can be written by using the digits 1 and 2

A
¹⁰C₁ + ⁹C₂

B
2¹⁰

C
¹⁰C₂

D
10!

Solution

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 .
Question 10
3 / -1

In a room there are 2 green chairs, 3 yellow chairs and 4 blue chairs. In how many ways can Raj choose 3 chairs so that at least one yellow chair is included?

A
3

B
30

C
64

D
84

Solution
Question 11
3 / -1

Find the number of non-congruent rectangles that can be found on a chessboard normal 8 x 8 chessboard,

A
24

B
36

C
48

D
None of these

Solution

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
Question 12
3 / -1

Find the number of integral solutions of equation
x + y + z + t = 29, x > 0 , y > l , z> 2 and t > 0.

A
²⁷C₃

B
²⁸C₃

C
2600

D
²⁹C₄

Solution

This situation is similar to distributing 29 identical things among 4 persons x, y, z and t in such a way that x is getting a minimum 1 thing, y is getting minimum 2 things, z is getting minimum 3 things and t is getting minimum 0 thing.
So, let us distribute 1 thing to x, 2 things to y, 3 things to z and 0 thing to t. After doing this distribution, we are free to distribute the remaining things among x, y, z and t in any ways.
Now the equation is:x + y + z + t = 23 The total number of ways of distributing 23 identical things among 4 persons in such a way that anybody can get any number of things = ²⁶C₃
Question 13
3 / -1

There is a regular decagon. Triangles are formed by joining the vertices of the polygon. What is the number of triangles which have no side common with any of the sides of the polygon?

A
50

B
300

C
44

D
294

Solution

. 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.
Question 14
3 / -1

The number of positive integral solution of abc = 30 is:

A
24

B
81

C
27

D
None of these

Solution

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
Question 15
3 / -1

A dinner menu is to be designed out of 5 different starters, 6 identical main courses and 4 distinct desserts. In how many ways menu be designed such that there is atleast one of each of the starters, main courses and desserts?

A
31 x 6 x 15

B
32 x 6 x 16

C
31 x 7 x 15

D
5 x 6 x 4

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Permutation And Combination Test - 3

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Permutation And Combination Test - 3

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Question 1

10 (9!)

8 (9!)

9 (9!)

None of these

Solution

Question 2

220

336

44

528

Solution

Question 3

12

10

60

None of these

Solution

Question 4

91

455

17C15

None of these

Solution

Question 5

890

1420

2700

3140

Solution

Question 6

6

18

12

9

Solution

Question 7

56

126

91

None of these

Solution

Question 8

15

12

10

18

Solution

Question 9

10C1 + 9C2

210

10C2

10!

Solution

Question 10

3

30

64

84

Solution

Question 11

24

36

48

None of these

Solution

Question 12

27C3

28C3

2600

29C4

Solution

¹⁷C₁₅

¹⁰C₁ + ⁹C₂

2¹⁰

¹⁰C₂

²⁷C₃

²⁸C₃

²⁹C₄