Division and Distribution of Objects into Groups — Formulas and Tricks

∴ Total number of ways of division = ¹⁰C₃ × 1 = ¹⁰C₇ × 1
=  \(\frac{{10!}}{{3!7!}}\)
now if we are also to distribute these divided objects into two distinct groups then there are 2! ways to distribute two distinct object into two person
Hence number of ways of distribution  = \(\frac{{10!}}{{3!7!}}\) × 2!

Example (2).

number of ways of Division = ¹⁰C₂ × ⁸C₃ × ⁵C₅
=  \(\frac{{10!}}{{2!8!}} \;\times\; \frac{{8!}}{{3!5!}}{\rm{ \;\times\; }}\frac{{5!}}{{5!0!}}\)
=  \(\frac{{10!}}{{2!3!5!}}\)

Now these divided packets can be arranged themselves into 3! ways.
Hence number of ways of Distribution = \(\frac{{10!}}{{2!3!5!}}\) × 3!

number of ways of division =  \(\frac{{4!}}{{2!2!}}\) =  \(\frac{{4 \times 3}}{2}\) = 6

BUT WAIT Let do it manually

Let’s take another example:-

Example:- Distribute 52 cards among 4 players equally.
Solution:-

(Q). Distribute 8 distinct objects into 3 distinct groups such that each group has at least 2 objects.
Solution:-

(Q.) We have to distribute n distinct objects among n children such that exactly one child get no onject.
Solution:-

(Q). In how many ways 8 toys can be distributed among three brothers such that such that each one receive at least one toy & no two get same number of toys & the youngest one receives the maximum number of toys.
Solution:-

(Q). In how many ways 10 distinct toys can be distributed among three brothers such that each one receive at least one toy & no two get same number of toys & the youngest one receives the maximum number of toys.
Solution:-

(Q). Distribute 5 different fruits among 3 bucket such that each bucket has at least one fruit.
Solution:-

(Q). A teacher has 31 students in his class. one day he decided to take all students for a picnic. So he called 3 cars & 4 auto rickshaws. He himself will go with his bike. In how many ways he can make them sit in vehicle, if seating capacity of car is 5 & that of auto rickshaw is 4. Internal arrangement within vehicle does not matter.
Solution:-

(Q). In the previous Example if 3 particular students insist on going by the same car then find the number of ways.
Solution:-

(Q). Distribute 5 different books among 3 students such that each student gets at least one book, also order of the books matters which a student has.
Solution:-

⇒ Actually in this question value of distinct object (5) & distinct groups (3) is less. So we can do it manually but if these values becomes more then it will become very difficult to do it manually. So in that case we will be required a formula to solve this type of problem quickly.
In Distribution of distinct object in distinct groups if arrangemet of objects within group is also important then:-
Number of ways to arrange n distinct object into r distinct group is:
                = n!.ⁿ ⁺ ʳ ⁻ ¹Cᵣ ₋ ₁    if blank group are allowed
                = n!ⁿ ⁻ ¹Cᵣ ₋ ₁         if blank group are not allowed i.e. each group must have at least one object

(Q). In the previous question:-
                  n = 5          r = 3

(i) when empty group not allowed
                = 5!.⁵⁻¹C₃₋₁ = 5!.⁴C₂ = 5!.6 = 720 Answer

(ii) when empty group alloweed
                = 5!.⁵⁺³⁻¹C₃₋₁ = 5!.⁷C₂ = 120×21 = 2520 Answer

(Q) In how many ways 10 different rings can be worn on 5 fingers if arrangement of rings in a finger matters.
(i) Any Finger has any number of rings.
(ii) Each Finger should have at least one ring.
Solution:-

n = 10
r = 5

(i) 10!.¹⁰⁺⁵⁻¹C₅₋₁ = 10!.¹⁴C₄ = \(10!.\frac{{14!}}{{10!4!}} = \frac{{14!}}{{4!}}\) Answer

(ii) 10!.¹⁰⁻¹C₅₋₁ = 10!.⁹C₄ Answer

(Q). In how many ways 5 distinct books can be distributed among 10 students such that each student gets at most 1 book.
Solution:-

(Q). In how many ways 5 distinct books can be distributed among 10 students such that each student gets at least 1 book.
Solution:- 

Since number of books (5) is less than number of students (10) & each student should get at least one book. So there is no such way.
Hence 0 Answer

Example:- Distribute 10 distinct object among 3 person in anyway.
Solution:- 

Here 1ˢᵗ object can be distributed in 3 ways.
Similarly 2ⁿᵈj object can be distributed in 3 ways.
3ʳᵈ object can be distributed in 3 ways
⋮
⋮
⋮
⋮
10ᵗʰ object can be distributed in 3 ways

Hence total number of ways of distribution
                  = 3×3×3×…………..10 times
                  = 3¹⁰ Answer

(Q). Distribute 100 letters in 10 letter boxes anyway.
Solution:-

1ˢᵗ letter can be distributed in 10 ways.
Similarly 2ⁿᵈ letter can be distributed in 10 ways.
3ʳᵈ letter can be distributed in 10 ways
⋮
⋮
⋮
⋮
100ᵗʰ letter can be distributed in 10 ways.

Hence total number of ways = 10×10×10×…………100 times
=10¹⁰⁰ Answer

(Q). A person has 5 servant at his home. He has 4 friends. He has to invite his friends by giving one letter to each. He tells his servants to invite his friends. In how many ways he can invite his friends.
Solution:-

S₁            F₁
S₂            F₂
S₃            F₃
S₄            F₄
S₅

which servant gives letter to friend will be a way.

∴ F₁ can be invited in 5 ways.
similarly F₂ can be invited in 5 ways
F₃ can be invited in 5 ways
F₄ can be invited in 5 ways

Hence total number of ways = 5×5×5×5
                                                    = 5⁴ Answer

(Q). Distribute 10 distinct books among 5 students if one particular person gets exactly 3 books.
Solution:- 

& Remaining 7 books will be distributed among remaining 4 students in anyway ⇒ number of ways sof doing this = 4⁷

Hence total number of ways = ¹⁰C₃×4⁷ Answer

(Q). Distribute 10 distinct books among 5 students if each one is entitled to get at most 9 books.
Solution:-

Required Answer = Total number of ways to distribute in any way – number of ways to distribute in which all 10 books are given to any single student
                                 = 5¹⁰ – 5 Answer

Example:- Distribute 10 identical chocolate among 3 students such that each student gets none or more chocolate.
Solution:-

This type of problem we have studied earlier in word formation when we were to find total number of words from the letters of the word
MISSISSIPPI
n = 11 (total number of letters)
p = 4 (S)
q = 4 (I)
r = 2 (P)

∴ total number of words =  \(\frac{{11!}}{{4!4!2!}}\)

Here also the same condition, where

∴ total ways = \(\frac{{12!}}{{10!2!}}\) Answer
This can be co-related with number of integral solution of equation.
In above question Let students be x, y & z
then

Here in place of ‘Lines(┃)’ we have used just ‘+’ symbol
To divide between 3 students we will be requiring 2 lines & in place of lines we have used ‘+’ sign.
∴ number of ways = number of solutions of above equation
                                  = ¹²C₂
                                  =  \(\frac{{12!}}{{2!10!}}\) Answer

(Q). In how many ways 5 identical chocolates can be distributed among 10 students if there is no restriction.
Solution:-

S₁ + S₂ + S₃ + S₄ + S₅ + S₆ + S₇ + S₈ + S9 + S₁₀ = 5
                                          S₁, S₂, S₃, ……. S₁₀ ≥ 0
total ‘+’ sign = 9
Hence number of ways = ¹⁴C₉ = \(\frac{{14!}}{{5!9!}}\) Answer

(Q). In how many ways 10 identical books can be distributed among 3 students if each student gets at least 1 book.
Solution:-

(Q). In how many ways 10 identical chocolates can be distributed among 4 children if a child has at most 1 chocolate.
Solution:-
                                0 Answer

(Q). In how many ways 4 identical chocolate can be distributed among 10 children such that a child gets at most 1 chocolate.
Solution:-

Simply we have to find number of ways of selecting 4 children out of 10 children.
= ¹⁰C₄ Answer

If chocolates are also 10 then number of ways = ¹⁰C₁₀ = 1 Answer

(Q). we have 30 identical pens. These are to be distributed among 4 students A, B, C, D such that student A gets at least 4 pens, student B gets none or more pens(Any nubmer of pen), student C gets at least 5 pens & student D gets at least 1 pen. Then in how many ways pen can be distributed.
Solution:-

now distribute 20 pens among A, B, C, D in anyway = ²³C₃ Answer

(Q). Find the number of terms in the expansion of (x + y + z)¹⁰⁰.
Solution:-

(x + y)² = x² + 2x¹y¹ + y²
               = x²y⁰ + 2x¹y¹ +x⁰ y²
i.e. sum of powers of variable x, y in every term = 2
Here total number of terms = number of ways to distribute power 2 i.e. 2 identical objects between 2 person x, y in anyway.

⇒(x + y)³ = x³ +3x²y¹ +3x¹y² + y³ 

Sum of power of variable x, y in every term = 3
∴ Number of terms = number of ways to distribute power 3 i.e. 3 identical object between 2 person x, y in any way.
∴ (x + y + z)¹⁰⁰ ⇒ Total numbers of terms
x + y + z = 100
                 & condition is x≥0
                                            y≥0
                                            z≥0
= ¹⁰²C₂ Answer

In this type of question wee will be required to find the value of Stirling Number of Second Kind.
Stirling Number of Second Kind:–
Formula for Stirling number of second kind is :-

= 63 + 3S(6, 2) + 9S(6, 3) + 4S(7, 3) + 16S(7, 4)
= 63 + 3(2⁶⁻¹ – 1) +9S(6, 3) + 4S(7, 3) + 16S(7, 4)
= 63 + 3×31 + 9[S(5, 2) + 3S(5, 3)] + 4[S(6, 2) + 3S(6, 3)] + 16[S(6, 3) + 4S(6, 4)]
= 156 + 9(2⁵⁻¹ – 1) +27S(5, 3) +4(2⁶⁻¹ – 1) + 12S(6, 3) + 16S(6, 3) + 64S(6, 4)
= 156 + 9×15 + 27S(5, 3) + 4×31 + 28S(6, 3) + 64S(6, 4)
= 415 + 27[S(4, 2) + 3S(4, 3)] + 28[S(5, 2) + 3S(5, 3)] + 64[S(5, 3) + 4S(5, 4)]
= 415 + 27[(2⁴⁻¹ – 1) + 3×⁴C₂] + 28[(2⁵⁻¹ – 1) + 3S(5, 3)] + 64[S(5, 3) + 4×⁵C₂]
= 415 + 27×7 + 81×6 + 28×15 + 84S(5, 3) + 64S(5, 3) + 64×4×10
= 4070 + 148S(5, 3)
= 4070 + 148[S(4, 2) + 3S(4, 3)]
= 4070 + 148[(2⁴⁻¹ – 1) + 3×⁴C₂]
 4070 + 148[7 + 3×6]
 = 4070 + 148×25
= 7770 Answer

Let us calculate the value of S(9, 4) using direct formula:-

Now come to question:-
(Q). In how many ways 7 people can be divided into 3 identical groups.
Solution:-

Since Groups are identical i.e. indistinguishable. So a particular object is in which group it does not matter. The only thing which matters is which objects are selected together.

∴ Total number of ways = 1 + 7 + 21 + 21 + 35 + 105 + 70 + 105
                                           =  365 ways Answer

(Q). In the previous question if the condition is that no group should be empty i.e. all groups should have at least one object.
Solution:-

So we need to consider only 4ᵗʰ, 6ᵗʰ, 7ᵗʰ & 8ᵗʰ cases.
which gives → 21 + 105 + 70 + 105 = 301 ways Answer

⦿ These were the manual way to do these type of questions. Now we will use Stirling number to solve these type of problems:-

(Q). (previous question); In how many ways 7 people can be distributed into 3 identical groups anyway.
Solution:-

\(\sum\limits_{i = 1}^3 {S(7,i)} \) = S(7, 1) + S(7, 2) + S(7, 3)
= 1 + (2⁷⁻¹ – 1) + S(7, 3)
= 64 + S(7, 3)
= 64 + \(\frac{1}{{3!}}\)[³C₀ 3⁷ – ³C₁ 2⁷ + ³C₂ 1⁷ ]
= 64 + \(\frac{1}{6}\)[ 1×2187 – 3×128 + 3×1 ]
= 64 +  \(\frac{1}{6}\)×1806
= 64 + 301
= 365 Answer

(Q). In how many ways 7 people can be distributed into 3 identical groups if each group gets at least 1 people(i.e. no group is empty).
Solution:-

S(7, 3) = 301 Answer

(Q). In how many ways 5 distinct balls can be distributed to 10 identical bags:-
(i) without any restriction
(ii) each bag gets atleast 1 ball
(iii) each bag gets at most 1 ball.
Solution:-

(ii) 0 Answer

(iii) 1 Answer

To solve these type of problems knowledge of generating functions will be required.

(Q). In how many ways 5 identical balls can be distributed to 10 identical bags if:
(i) There is no restriction
(ii) Each bag must contain atleast 1 ball
(iii) Each bag must contain at most 1 ball
(iv) Each bag must contain exactly 1 ball
Solution:-

Let first do it manually:-