## Mathematics MCQ on Permutations and Combinations for JEE and Engineering Exam 2022

MCQ on Permutations and Combinations: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.

In this post we are providing you MCQ on Permutations and Combinations which will be beneficial for you in upcoming JEE and Engineering Exams.

## MCQ on Permutations and Combinations

Q1. In how many ways can 15 members of a council sit around a circular table, when the Secretary is to sit on one side of the Chairman and the Deputy Secretary on the other side?
a) 10 ! ×2
b) 12 ! ×2
c) 13 ! ×2
d) 15 ! ×2

Since the total members are 15, but one is to the left, because of circular condition. Therefore, the remaining members are 14, but three special members constitute a member. Therefore the required number of arrangements are 12 ! ×2, because the chairman remains between the two specified persons and the person can sit in two ways.

Q2. 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) 876
b) 879
c) 877
d) 875

p = 10, q = 9, r = 7

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

= [11 * 10 * 8] – 1

= 879

Q3. In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together?
a)  5 ! ×5 !
b) 4 ! ×5 !
c) 3 ! ×5 !
d) 4 ! ×6 !

Since the total number of ways in which boys can occupy any place is (5 − 1)! = 4 ! and the 5 girls can sit accordingly in 5 ! ways. Hence the required number of ways is 4 ! ×5 !.

Q4. Number of divisors of n = 38808 (except 1 and n) is _____.
a) 65
b) 72
c) 70
d) 68

Since, 38808 = 8 × 4851

= 8 × 9 × 539

= 8 × 9 × 7 × 7 × 11

= 2× 32 × 72 × 11

So, the number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 72.

This includes two divisors 1 and 38808.

Hence, the required number of divisors = 72 – 2 = 70.

Q5. The total number of positive integral solution for x, y, z such that x* y * z = 24, is ________.
a) 36
b) 32
c) 34
d) 30

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

Q6. 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)  328
b) 325
c) 320
d) 320

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.

Q7. An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is
a) 6
b) 5
c) 7
d) 2

Since at any place, any of the digits 2, 5 and 7 can be used, the total number of such positive n-digit numbers are 3n. Since we have to form 900 distinct numbers, hence 3≥ 900 ⇒ n = 7.

Q8.  From 6 different novels and 3 different dictionaries, 4 novels and a dictionary are to be selected and arranged in a row on the shelf such that the dictionary is in the middle. What is the number of such arrangements?
a) 1080
b) 1088
c) 1085
d) 1082

4 novels can be selected from 6 novels in 6C4 ways. 1 dictionary can be selected from 3 dictionaries in 3C1 ways. As the dictionary selected is fixed in the middle, the remaining 4 novels can be arranged in 4! ways.

Hence, the required number of ways of arrangement = 6C4 * 3C1.* 4! = 1080

Q9. A five-digit number divisible by 3 has to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is ________.
a) 214
b) 215
c) 216
d) 217

We know that a five-digit number is divisible by 3, if and only if the sum of its digits (= 15) is divisible by 3, therefore we should not use 0 or 3 while forming the five-digit numbers.

Now,

(i) In this case, we do not use 0; the five-digit number can be formed (from the digit 1, 2, 3, 4, 5) in 5Pways.

(ii) In this case, we do not use 3; the five-digit number can be formed (from the digit 0, 1, 2, 4, 5) in 6 ! −5! × 2 = 480 ways.

The total number of such 5 digit number = 5P+ (5P− 4P4)

= 120 + 96

= 216.

Q10. 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) 999
b) 997
c) 899
d) 888

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.

By Team Learning Mantras

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