*Thursday, July 23rd, 2020*

Division and distribution is an important topic in permutation and combination. In this topic we learn how to divide the objects in groups or how to distribute these objects between different persons. This topic contains lots of categories. First category is we have to look which type of object we have to distribute either **identical** or **distinct**. Here we will discuss the concept of distribution of distinct objects later in second article we will discuss distribution of identical object. Letâ€™s start to understand this concept with various examples.

Letâ€™s take an to understand this concept let us assume we have to divide 5 different balls in two groups in such a way that two balls are in one group and remaining three are in another group. Now think about it if you select any two balls for your first group then the remaining three are the balls of other group or vice- versa. So the number of ways of selecting two balls out of five balls is

^{5}C_{2} i.e. 5!/(2! 3!).

Now if we have to distribute these groups in two persons the question may be asked as follows:

In how many ways five different balls can be distributed in two persons such that one person get two balls and other person get remaining three balls. Now try to understand the difference between this concept and the concept we discussed above. In above concept is we make a group of two balls the another group of three balls is automatically formed or vice versa in that case it was not important that which group we made either we selected two balls or three. But here the order of the group is also important means now this is also important that who is getting a group of two balls and who is getting a group of three balls. To understand this first we pack the balls in two packets such that one packet contains two balls and another contains three balls in ^{5}C_{2 }after that we will distribute these packets between two persons in 2! ways. So in this counting we have to multiply the number of ways of division by 2! also as in this case it is important that who is getting a packet of two balls and who is getting other packet.

**Â ****Eg1: **In how many ways 10 different objects can be divided in 3 groups of group sizes of two three and five objects?

**Â ****Solution: **Now it is clear from the language of question that we just divide the objects in three packets we donâ€™t have to distribute these packets to anyone.

First, we select two objects out of ten objects in ^{10}C_{2} ways, then we make another packet of three objects in ^{8}C_{3} ways and the last packet is automatically formed of 5 objects means the number of ways of making last packet is only one.

So the answer of above question is first we make a group of two objects in ^{10}C_{2} ways then we make another group of 3 objects in ^{8}C_{3 }ways then the group of 5 objects can be made by only one way. Total number of ways is ^{10}C_{2 }Ã— ^{8}C_{3} Ã— 1

10!/2!8!Ã—8!/3!5!Ã—1 i.e. (10!/2!3!5!)

**Eg2: **In how many ways 10 different objects can be distributed in 3 persons in such a way that they get two, three and five objects.

**Â ****Solution: **In this case first we make three packets of sizes two, three and five objects in ^{10}C_{2 }Ã— ^{8}C_{3} Ã— 1 ways as discussed above then we multiply this counting by 3! as we have to distribute these packets in three persons also, so it is important to consider that who is getting a packet of 2 objects, 3 objects and 5 objects. So the answer of above question is (10!/2!3!5!)3!

In general we can say that if we have p + q + r objects and we have to divide them in three groups of p, q, and r objects, so the number of ways will beÂ ^{p+q+r}C_{p }Ã— ^{q+r}C_{q} Ã— ^{r}C_{rÂ }i.e.Â (p+q+r!)/p!q!r! (when order of the group is not considered.) and

((p+q+r!)/p!q!r!)3! (When order of the group is also considered).

In case of distribution first we calculate number of ways of division and then multiply it by n! if we have to distribute these groups in n persons.

**Eg3:** In how many ways 4 objects can be divided in two groups such that each group contains 2 objects each.

**Solution:** According to the above discussion the number of ways of division is 4! Will be in numerator and two times 2! will be in denominator as we made two groups of group size 2 objects. So total number of ways will be (4!/2!2!) = 6 **(but in this case this is not a correct answer.) **

Letâ€™s take an example there are 4 objects a, b, c and d and we have to divide them in two groups each group contains two objects. Letâ€™s start counting manually

ab | cd |

ac | bd |

ad | bc |

bc | ad |

bd | ac |

cd | ab |

In above table you can see that counting contains a case multiple time means if we make a group of ab another group of cd is automatically formed and vice versa this case cannot be counted twice as there is no difference between both the cases. **So correct answer is (4!/2!2!2!) = 3.Â **

Means we can say that if same group size is repeated n times we have to divide additionally our answer by n!

**Eg4: **In how many ways twelve objects can be divided in five groups such that three groups contains two objects and remaining two groups contains three objects**?**

**Solution: Â **We have to make five groups so the number of ways will beÂ

We have additionally divided our answer with 3! and 2! as group size of 2 objects is repeating 3 and group size of 3 objects is repeating 2 times respectively.

**Eg5: **In how many ways divide 3n objects

(a)Â Â Â Â Â In three groups such that each group contains n objects

(b)Â Â Â Â Â in three persons such that each person gets n objects

**Solution:**

**Case a)** In this case we have to divide the objects in 3 groups only but see carefully that the size is same of all three groups so we have to divide number of ways of division by 3! also.Â So total number of ways is

**Case b) **In this case after division in groups we have to distribute in 3 persons also means we have to multiply the answer of case a with 3! Also. So total number of ways in dividing three persons is

**Â ****Eg6: **In how many ways eleven distinct objects can be divided in three groups such that each group contains at least three objects?

**Â ****Solution: **In this question first we make three packets and keep three objects in each packet now we have two more object so there are two cases

**Case 1:** If we keep both objects in one of the packets so the group size will be (5, 3, 3), so in this case 11! will be in numerator as we have to divide 11 objects in three groups and 5! and 3! two times will be in denominator as there are two groups of same size so there is 2! also in denominator so total number of ways of this case will be

**Case 2:** If we divide both objects in any two of the packets so group size will be (3, 4, 4). so in this case 11! will be in numerator as we have to divide 11 objects in three groups and 3! and 4! two times will be in denominator as there are two groups of same size so there is 2! Also in denominator so total number of ways of this case will be

So both the cases are mutually exclusive then total number of ways will be

Letâ€™s take a different example to understand this concept with more clarity.

**Â ****Eg7: **If there are five elements in set A and three elements in set B then find how many onto functions are possible?

**Solution: Â **In the first look, the student thinks that this question is based on function and graphs but we can understand this counting simply with division and distribution concept. A function is on-to function if for every element b in the co-domain B of *f* there is at least one element a in the domain A. Let us assume five elements in set A are (a_{1, }a_{2, }a_{3, }a_{4,} a_{5}) and three elements in set B are (b_{1}, b_{2, }b_{3}). So we can consider this question as supposing we have 5 different objects and we have to distribute these objects between three persons such that each person gets at least one object. In that case first we make three packets and put one object in each packet now for remaining two objects we have two cases as discussed above either both the object are in one of the packets to the group size will be (3, 1, 1) **or **we divide both objects in any two of the packets to the group size will be (1, 2, 2). So the number of ways of distribution or number of onto functions is

i.e. 60 + 90 = 150. So we simply understand the counting of on-to function with the concept of division and distribution.

As you can see the topic is quite wide and tricky also, you will face similar problems in the exam where you have to decide the approach of counting. You will get more idea of the type of questions you need to practice through past year CAT papers. This is a very important topic in permutation and combination. Especially the last example is quite tricky and we have complete knowledge of function and graph and permutation and combination to attempt that question. Hopefully, I am able to explain the concept clearly.

Permutation and Combination â€“ Distribution of Objects

Permutation and Combination â€“ Fundamental Principle of Counting

Set Theory- Maximum and Minimum Values

How to solve questions based on At least n in Set Theory for CAT Exam?

Permutation and Combination â€“ 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

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),â€¦â€¦, then a1 + a2 +â€¦â€¦..+ a100 is

Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,â€¦ has the property that an = 3(a(n+ l) + a(n+2) +â€¦.) for every n â‰¥ 1. If the sum a1 + a2 + a3 +â€¦â€¦. = 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,â€¦â€¦..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + â€¦.+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + â€¦. + 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

Use referral code **HANDA** to get 10% off.

## Leave a Reply