JEE Main Mathematics Permutations and Combinations is an important chapter and each year, questions from this chapter are asked in the exam. Questions asked from this chapter are mostly scoring in case you are clear with the concepts. Permutation is complex in comparison to Combination. Almost 1-2 questions are asked in JEE Main from the chapter- Permutation and Combination, which together holds a combined weightage of 4 marks.
Engineering candidates giving JEE Main can solve any question, in case they have basic understanding of the concepts. Beneath we have presented a collection of JEE Main Permutations and Combinations past year questions with solutions so that candidates can practice from the most reliable tool. Check JEE Main Revised Exam Pattern
JEE Main - Previous Year Asked Questions and Solutions
Question 1: A letter lock that has three rings is marked differently with ten letters. In how many ways is it possible to make an unsuccessful attempt to open the lock?
Solution:
Two rings might have the same letter at a time, but the same ring will not have two letters at the same time. Thus we must proceed ring wise. Each of the three rings will have any one of the 10 different letters in 10 ways.
Thus the total attempts would be= 10 × 10 × 10 = 1000. However, out of these 1000 attempts, only one attempt would be successful.
Therefore, the needed number of unsuccessful attempts would be 1000 less 1 = 999.
Question 2: The total number of positive integral solution for x, y, z such that x* y * z = 24 will be?
Solution:
We have,
x* y * z = 24
x* y * z = 23 × 31
Total methods of distributing ‘n’ identical balls into ‘r’ different boxes is (n + r − 1)C(r − 1)
We are supposed 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
Question 3: What will be the total number of signals that can be made using five flags of a different color, when any number of them can be used in any signal.
Solution:
Case I: When just one flag is used. Number of signals made would be= 5P1 = 5.
Case II: When two flags are used. Total signals made = 5P2 = 5 * 4 = 20.
Case III: When only three flags are used. Total signals made would be = 5P3 = 5 * 4 * 3 = 60.
Case IV: When four flags are used. Total signals made = 5P4 = 5 * 4 * 3 * 2 = 120.
Case V : When five flags are used. Number of signals made = 5P5 = 5! = 120.
Hence, the total number would be 325. 5 + 20 + 60 + 120 + 120 = 325.
Question 4: You need to prove that if each of the ‘m’ points in one straight line is joined to every n points on the other straight line, except the points on the given two lines. The number of points of intersection of these lines is \frac{1}{4}4 mn (m-1 ) (n-1).
Solution:
To get one point of intersection, there is a need of two points on the first line and two points on the second line. We can select them out of n-points in nC2 ways and for m points in mC2 ways.
Therefore, the needed number would be mC2 × nC2 = (m(m-1))/2! x (n(n-1))/2! = \frac{1}{4}4 m n (m – 1)(n – 1)
Question 5: The letters of the word SACHIN are arranged in maximum possible ways, and these words are written in a dictionary. Count the serial number at which the word SACHIN will come?
Solution:
The correct alphabetical order is A, C, H, I, N, S
Total words starting with A – 5!
Total 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.
Question 6: The value of 50C4 + \sum_{r=1}^{6}∑r=16
56-rC3 is ________.
Solution:
Question 7: Let’s assume that all the balls are similar excluding the difference in colours. What will be the total methods in which one or more balls can be chosen from 10 white, 9 green and 7 black balls?
Solution:
p = 10, q = 9, while r would be equal to \7
Different methods to select = (p + 1) * (q + 1) * (r + 1) – 1
= [11 * 10 * 8] – 1
= 879
Question 8: From a total of 6 different novels and 3 different dictionaries, 4 novels and a dictionary needs to be chosen and arranged in a row on the shelf such that the dictionary is in the middle. What is the total number of these arrangements?
Solution:
4 novels can be chosen from 6 novels in 6C4 ways. 1 dictionary can be chosen from 3 dictionaries in 3C1 ways. As the dictionary selected is fixed in the middle, the left out 4 novels will be arranged in 4 ways.
Hence, the total methods of arrangement = 6C4 * 3C1.* 4! = 1080
Question 9. Of the total ways of selecting n cards out of an unlimited number of cards containing the numbers 0, 2, 3, such that it cannot be utilized to mention the number 203 is 93, then n would be equivalent to -
(a) 3
(b) 4
(c) 5
(d) 6
Solution:
203 will not be written if in the choice of n cards we get either (2 or 3), (3 or 0), ( 0 or 2), (only 0), (only 2) or (only 3).
JEE Main Permutations and Combinations Quick Formulas
Importance of learning formulas is not newer for the JEE aspirants, having a clear understanding of the formulas help candidates in getting the answer quickly. Some of the formulas that one must pen down in notepad so that they can revise them later include:
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In Mathematics, the factorial is expressed as ‘!’. If we have to show 5 factorials, then we will write it as 5! Typically, the factorial of any positive number n will be mentioned as n!.
Mathematically,
So, 4! = 3! x 4 = 2! x 3 x 4 = 1! x 2 x 3 x 4 = 0! x 1 x 2 x 3 x 4 = 1 x 2 x 3 x 4
In the similar methods, for any positive integer ‘n’ would be,
n! = n x (n – 1) x (n – 2) x …….. x 3 x 2 x 1.
Therefore to conclude with we can say that factorial of any positive integer ‘n’ as ‘the product of all the positive integers would be less than or equal to n’.
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Permutation = Selection x Arrangement
This can be represented by the mathematical relation.
We know that,
he number of permutations or arrangement of n distinct things taken all at once can be expressed as -
Theorems on Combination
The combination of n distinct objects selected r at a time will be written and calculated as -
JEE Main aspirants can refer to these Permutations and Combinations previous year questions with solutions to ace their preparations. This set of questions will help them in understanding the kind of questions asked in the exam.
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