# Permutation & Combination Concepts by Gaurav Sharma - Part (2/3)

• Direct Method : Number of ways to distribute n identical things to r persons such that anyone can get any number of things is (n+r-1)C(r-1)

Example : Number of ways to distribute 15 ladoos among 5 persons is (15 + 5 – 1) C (5-1) = 19C4

If we have to distribute 15 ladoos among 5 persons such that all of them get atleast one ladoo, then
we have the equation: A + B + C + D + E = 15
Now, give one ladoo to each one of them so put A = a + 1, B = b + 1, C = c + 1, D = d + 1 and E = e + 1
We get, a + b + c + d + e = 10
This case is the same as the case where we have to distribute 500 ladoos among 26 persons such that each person can get any number of ladoos.
So, the number of solutions are (n + r – 1 ) C (r -1 )
N = 10 and r = 5, so (10 + 5 – 1) C ( 5 – 1) = 14C4

Direct method : Number of ways of distributing n things to r persons such that all get atleast one are: (n – 1)C(r – 1)

Positive integral solutions: Where the variables can be any natural number
Non negative integral solution: Where the variables can be any whole number
Integral solution: where the variable can be any integer

Find the number of positive integral solutions of a + b + c + d = 500?

We have to find the POSITIVE integral solutions of a + b + c + d = 500 which means we have to find the natural number of solutions for the equation.

This case is the same as the case where we have to distribute 500 ladoos among 4 persons such that each gets atleast 1 ladoo.

So the number of solutions are (n – 1) C (r – 1)

Here n = 500 and r = 4

So, (500 – 1) C (4 – 1) = 499 C 3

Find the non negative integral solutions of a + b + c + d + e .. + z = 500

We have to find non negative integral solutions for the equations a + b + c + d + .. + z = 500, which means we have to find the whole number solutions for the equation.

This case is the same as the case where we have to distribute 500 ladoos among 26 persons such that each person can get any number of ladoos.

So, the number of solutions are (n + r – 1 ) C (r -1 )

Here, n = 500 and r = 26

So, (500 + 26 – 1)C(26 – 1) = 525C25

Find the number of even natural solutions of a + b + c = 100

a + b + c = 100

So, put a = 2x, b = 2y and c = 2z

So 2x + 2y + 2z = 100

x + y + z = 50

We need to find even natural solutions of the equation x + y + z = 50

So the number of solutions = (50 – 1)C(3 – 1) = 49C2

Find the number of even non negative solutions of a + b + c + d + e = 200

Put a = 2x, b = 2y and c = 2z, d = 2w and e = 2v

2x + 2y + 2z + 2w + 2v = 100

x + y + z + w + v = 100

So the number of solutions = 100 + 5 – 1 C 5 – 1 = 104C4

Find the odd non negative integral solutions of a + b + c = 159

Put a = 2x + 1, b = 2y + 1 and c = 2z + 1

2x + 2y + 2z + 3 = 159

2 ( x + y + z) = 156

x + y + z = 78

Where x, y and z can take values which belong to whole numbers.

So the number of solutions: (n + r – 1)C(r – 1)

n = 78 and r = 3

So, ( 78 + 3 – 1)C(3-1) = 80C2

Find the number of odd solutions of a + b + c + d = 100

Put a = 2x + 1, b = 2y + 1, c = 2z + 1 and d = 2v + 1

2x + 2y + 2z + 2v + 4 = 100

2 (x + y + z + v) = 96

x + y + z + v = 48

Where x, y, z and v can take values which belong to whole numbers because when x = 0 we get a = 1, which is natural number. So the number of solutions: (n + r – 1)C(r – 1)

Where n = 48 and r =4

So (48 + 4 – 1)C(4 -1) = 51C3

Find the number of odd natural solutions of a + b + c + d + e = 200

Odd numbers added even number of times = Even number
Odd numbers added odd number of times = Odd number

So number of solutions for this equation is Zero.

Find the number of integral solutions of a + b + c + d = 50 such that each variable is greater than 5

Put a = A + 5, b = B + 5, c = C + 5 and d = D + 5

We get A + B + C + D = 50 – 5 x 4 = 30

So, now we have to find positive solutions for

A + B + C + D = 30

Which is (n – 1)C(r – 1), where n = 30 and r = 4

We get, (30 – 1)C(4 – 1) = 29C3

Find the number of positive solutions of a + b + c = 40 such that they all are distinct

We need to find Total Solutions – ( Number of solutions where any 2 of the variables are equal + Number of solutions where all 3 variables are equal)

Total positive solutions = (40 – 1)C(3 – 1) = 39C2

Number of solutions where any 2 of the variables are equal:

If a = b we get 2a + c = 40 Where a can take any value from 1 – 19. So, 19 cases.

If a = 20 then c = 0 which is not considered as a positive solution. Similarly when b = c ( 19 cases ) and c = a ( 19 cases )

And all three of them cannot be equal because 40 is not a multiple of 3

So distinct solutions = 39C2 – ( 19 x 3 ) = 684

Find the number of non-negative integral solutions of a + b + c = 28 such that they all are distinct

We need to find Total solutions – ( Number of solutions where any 2 of the variables are equal + Number of solutions where all three are equal )

Total positive solutions = (28 + 3 – 1) C (3 – 1) = 30C2

Number of the solutions where any 2 of the variables are equal:

If a = b, we get 2a + c = 28, where a can take any value from 0 – 14. so 15 cases.

If a = 14 then c = 0 which is considered a non-negative solution.

Similarly when b = c ( 15 cases) and c = a (15 cases)

And all three of them cannot be equal because 28 is not a multiple of 3

So distinct solutions = 30C2 – (15 x 3) = 390

How many terms are in the expansion of (a + b + c + d)^20

Sum of powers of each variable in the expansion of (a+b+c+d)^20 will be 20.

We have to find the number of ways in which 20 can be divided among 4 variables

So non negative solutions of a + b + c + d = 20

(n + r – 1)C(r – 1) where n = 20 and r = 4

S0 ( 20 + 4 – 1)C(4 – 1) = 23C3 = 1771

Number of terms in the expansion of (a + b + c + d)^20 = 1771

Find the number of ways of distributing 27 ladoos to Swetabh, Raman and Gaurav such that number of ladoos Swetabg gets are more than the number of ladoos Raman gets which in turn is more than the number of ladoos Gaurav gets

S + R + G = 27

S > R > G

So S, R and G are distinct

So total solutions – (cases where 2 are equal + cases where all 3 are equal)

Now total non-negative solutions: (27 + 3 – 1)C(3 – 1) = 29C2

Number of solutions where any 2 of the variables are equal: If S = R, we get 2S + R = 27 where S can take any value from 0 – 13. So, 14 cases but this includes cases where S = R = G, so 14 – 1 = 13 cases.

Similarly when R=G (13 cases) and when S = G (13 cases)

And 1 case where S = R = G = 9

So distinct solutions = 29C2 – ( 13 x 3 ) – 1 = 366

And for cases where S > R > G we divide 366 by 3! (Because we want only one of the 6 arrangements possible)

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