Maths Quiz on Permutations and Combinations for JEE Mains/Advance

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Maths Quiz on Permutations and Combinations
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JEE Maths Quiz on Permutations and Combinations

JEE Maths Quiz on Permutations and Combinations : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Permutations and Combinations.

Quiz on Permutations and Combinations

 

Q1. Find the total number of signals that can be made by five flags of a different color when any number of them may be used in any signal.
a) 325
b) 575
c) 850
d) 225

View Answer

a) 325

Solution:

Case I: When only one flag is used. No. of signals made = 5P1 = 5.

Case II: When only two flags are used. Number of signals made = 5P2 = 5 * 4 = 20.

Case III: When only three flags are used. Number of signals is made = 5P3 = 5 * 4 * 3 = 60.

Case IV : When only four flags are used. Number of signals made = 5P4 = 5 * 4 * 3 * 2 = 120.

Case V : When five flags are used. Number of signals made = 5P5 = 5! = 120.

Hence, required number = 5 + 20 + 60 + 120 + 120 = 325.

Q2. If the letters of the word SACHIN are arranged in all possible ways, and these words are written in a dictionary, at what serial number does the word SACHIN appear?
a) 601
b) 596
c) 482
d) 976

View Answer

a) 601

Solution:

(1) Alphabetical order is A, C, H, I, N, S

No. of words starting with A – 5!

No. of words starting with C – 5!

No. of words starting with H – 5!

No. of words starting with I – 5!

No. of words starting with N – 5!

SACHIN-1

601.

Q3. Assuming the balls to be identical except for the difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is ______.
a) 499
b) 572
c) 879
d) 786

View Answer

c) 879

Solution:

p = 10, q = 9, r = 7

Total ways of selection = (p + 1) * (q + 1) * (r + 1) – 1

= [11 * 10 * 8] – 1

= 879

Q4. A letter lock consists of three rings, each marked with ten different letters. In how many ways is it possible to make an unsuccessful attempt to open the lock?
a) 100
b) 99
c) 999
d) 1000

View Answer

c) 999

Solution:

Two rings may have the same letter at a time, but the same ring cannot have two letters at a time. Therefore, we must proceed ring wise. Each of the three rings can have any one of the 10 different letters in 10 ways.

Therefore, the total number of attempts = 10 × 10 × 10 = 1000.

But out of these 1000 attempts, only one attempt is successful.

Therefore, the required number of unsuccessful attempts = 1000 – 1 = 999.

Q5. The total number of positive integral solution for x, y, z such that x* y * z = 24, is ________.
a) 20
b) 30
c) 40 
d) 50

View Answer

b) 30

Solution:

We have,

x* y * z = 24

x* y * z = 23 × 31

The number of ways of distributing ‘n’ identical balls into ‘r’ different boxes is (n + r − 1)C(r − 1)

Here we have to group 4 numbers into three groups

Number of integral positive solutions

(3 + 3 − 1)C(3 − 1) × (1+ 3 − 1)C(3 − 1)

5C2 × 3C2

= 30

 


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