Permutation and Combination – Distribution of Objects

Tuesday, October 16th, 2018


Permutation and Combination - Distribution of Objects

As an astute man Mr. Gump once said, “Life is like a box of chocolates. You never know what you going to get.” The Permutations and Combinations that life presents us daily is baffling and probably it is because of that inherent fear of choices and cases we get intimidated by such questions in the exam. I understand that Permutation and Combination is one of the dreaded topics but I hope that once you understand the fundas given below, your fear will reduce.

Permutation and Combination Funda 1: De-arrangement

If ‘n’ distinct items are arranged in a row, then the number of ways they can be rearranged such that none of them occupies its original position is:

1

Note: De-arrangement of 1 object is not possible.

Dearr(2) = 1; Dearr(3) = 2; Dearr(4) =12 – 4 + 1 = 9; Dearr(5) = 60 – 20 + 5 – 1 = 44

Eg1.1: A person has eight letters and eight addressed envelopes corresponding to those letters. In how many ways can he put the letters in the envelopes such that exactly 5 of them get delivered correctly?

Solution: At first, select the five letters that get delivered correctly. That can be done in 8C5 ways.

Now, the other three must get delivered to the wrong address. That can be done in Dearr(3) = 2 ways.

So, total ways is 2 x 8C5 = 2 x 56 = 112 ways.

Permutation and Combination Funda 2: Partitioning

‘n’ identical items in ‘r’ distinct groups No restrictions:           n+r-1Cr-1
No group empty:         n-1Cr-1
‘n’ distinct objects in ‘r’ distinct groups No restrictions:            rn
Arrangement in a group is important:

Note: Other than standard distribution / partitioning problems, these ideas can be used to solve questions in which number of solutions are asked.

Eg 2.1: How many solutions are there to the equation a + b + c = 100; given that

a)      a, b and c are whole numbers.

b)      a, b and c are natural numbers.

Solution:

Case a) is identical to a case in which 100 identical chocolates are being distributed in three kids a, b and c. It is possible that one kid gets all the chocolates. In this case, we will use the formula for distributing ‘n’ identical items in ‘r’ distinct groups where n = 100 and r = 3.

So, it can be done in 102C2 ways.

Case b) is identical to a case in which 100 identical chocolates are being distributed in three kids a, b and c. Every kid must get at least one chocolate. In this case, we will use the formula for distributing ‘n’ identical items in ‘r’ distinct groups where no group is empty and n = 100 and r = 3.

So, it can be done in 99C2 ways.

Eg 2.2: In how many ways can you distribute 5 rings in

a)      4 boxes.

b)      4 fingers.

Solution: First of all we need to identify the difference between distributing in boxes and distributing in 4 fingers. The distinction is that in case of fingers, unlike boxes, the order in which rings are placed matters.

In Case a; Ring 1 can go in any of the four boxes, so it has four choices. Ring 2 can also go in any of the four boxes, so it has four choices. Similarly for Ring 3, Ring 4 and Ring 5; there are 4 choices each. So, the total number of ways of distribution is = 4 x 4 x 4 x 4 x 4 = 45. This is essentially how the formula rn is derived.

In Case b; Ring 1 can go in any of the four fingers, so it has 4 choices.

Ring 2 can go in any of the four fingers but it has five choices. There is a finger, say F3, which contains the ring R1. Now, on F3, R2 has two choices – it can go above R1 or below R1. So, the total number of choices for R2 is 5.

Ring 3 can go in any of the four fingers but it now has 6 choices.

Ring 4 can go in any of the four fingers but it will now have 7 choices.

Ring 5 can go in any of the four fingers but it will now have 8 choices.

So, the total number of way of distribution of rings is =

3

This is essentially how the

4formula  is derived.

Permutation and Combination Funda 3:

Number of ways of arranging ‘n’ items, out of which ‘p’ are alike, ‘q’ are alike and ‘r’ are alike given that p + q + r = n

Number of ways of distributing ‘n’ distinct items, in groups of size ‘p’, ‘q’ and ‘r’ given that p + q + r = n

5

I hope that this would help you solve problems in the exam. May be the chocolate you end up getting is a Bournville. May be you would have earned it.

Other posts related to Quantitative Aptitude – Modern Maths

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How to solve questions based on At least n in Set Theory for CAT Exam?
Permutation and Combination – Fundamental Principle of Counting
How to find Rank of a Word in Dictionary (With or Without Repetition)
Basic Probability Concepts for CAT Preparation
Sequence and Series Problems and Concepts for CAT Exam Preparation

CAT Questions related to Quantitative Aptitude – Modern Maths

All questions from CAT Exam Quantitative Aptitude – Modern Maths
Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q4: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q5: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?
Quantitative Aptitude – Modern Maths – P&C – Q5

You can also see Permutation and Combination – Fundamental Principle of Counting

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Permutation and Combination – Distribution of Objects

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