Non-Negative Integers include integers {0, 1, 2, 3, 4, ....}

a + b + c = 10

METHOD-1
at a = 0, b+c = 10 i.e. (b, c) may be (0, 10) (1, 9) (2, 8) (3, 7) (4, 6) (5, 5) (6, 4) (7, 3) (8, 2) (9, 1) (10, 0) i.e. 11 solutions

at a = 1, b+c = 9 i.e. (b, c) may be (0, 9) (1, 8) (2, 7) (3, 6) (4, 5) (5, 4) (6, 3) (7, 2) (8, 1) (9, 0) i.e. 10 solutions

ie. we notice a series of 11+10+9+8+7+6+5+4+3+2+1 = (1/2)*11*12 = 66

METHOD-2

Whole number solution of an equation is given by \((n+r-1)C_{(r-1)}\)

here n = 10, r = 3 (number of variables)

so total whole number solutions = (10+3-1)C(3-1) = 12C2 = 66

To know the derivation of this property, please refer teh video



Answer: Option C
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a+b+c=10

\(\frac{a+b+c}{3}\) = \(3.33\)

The highest among them can take 7 values(4, 5, 6, 7, 8, 9 and 10)

Case 1- The highest number is 10.

Possible triplet (a,b,c) = (10,0,0)
Total arrangements possible \(= \frac{3!}{2!} = 3\)

Case 2- The highest number is 9.

Possible triplet (a,b,c) = (9,1,0)
Total arrangements possible \(= 3! = 6\)


Case 3- The highest number is 8.

Possible triplets (a,b,c) = (8,1,1) and (8,2,0)
Total arrangements possible \(= \frac{3!}{2!}+3! = 9\)

Case 4- The highest number is 7.

Possible triplets (a,b,c) = (7,2,1) and (7,3,0)
Total arrangements possible \(= 3!+3! = 12\)

Case 5- The highest number is 6.

Possible triplets (a,b,c) = (6,4,0), (6,3,1) and (6,2,2)
Total arrangements possible \(=3!+ 3! +\frac{ 3!}{2!} = 15\)

Case 6- The highest number is 5.

Possible triplets (a,b,c) =(5,5,0), (5,4,1) and (5,3,2)
Total arrangements possible \(= \frac{ 3!}{2!}+3!+3! = 15\)

Case 7- The highest number is 4.

Possible triplets (a,b,c) = (4,4,2) and (4,3,3)
Total arrangements possible \(= \frac{3!}{2!} + \frac{3!}{2!} = 6\)

Number of ordered triplets (a, b, c), where a, b, and c are non-negative integers, are there such 'a + b + c = 10' =3+6+9+12+15+15+6 = 66

The answer should be C

a,b and c are non-negative => they can be from 0-10, since a+b+c = 10

Now, solving at length:
Step 1: a = 0=> b+c = 10, where b may start from 0(c=10) and continue till 10(c=0).
Step 2: counting such pairs of b and c = (0, 10), (1, 9) ,(2, 8) ,(3, 7) ,(4, 6) (5, 5) (6, 4) (7, 3) (8, 2) (9, 1) (10, 0) = 11 pairs
Moving on to a=1
Step 3: a = 1 => b+c = 9, where b may start from 0(c=9) and continue till 9(c=0).
Step 4: counting such pairs gives = (0, 9) (1, 8) (2, 7) (3, 6) (4, 5) (5, 4) (6, 3) (7, 2) (8, 1) (9, 0) = 10 pairs

If we continue doing this, we will see it goes like 11,10,9,.....1. The sum of which is 66.

This is the same as distributing 10 identical toffees among 3 children.

Let the toffees be denoted as T.
There are 10 Ts in a line.

To distribute these among 3 children, we need to make 2 partitions (shown as p in the diagram below):

T T T T p T T p T T T T

Depending on where we place the 'p', we will get different distributions for the 3 children.
However, as we keep doing this, we keep making different words.

Thus, the number of distributions is the same as the number of words formed.

Total words formed by 10 T and 2 p
= 12!/(10!*2!)
= 12*11/2 = 66

Answer C

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sujoykrdatta,

Right Sir, Thank you for your response

What I have understood is this.

For non negative integers we DO NOT need to distribute one to each and then do the partition and re arrange process.

For positive integers we need to distribute one to each and then do the partition and rearrange process.

is my understanding correct?
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Perfectly put.

Given this understanding, we can formulate these 2 things:

1. Number of positive integer solutions for r integers adding up to n
= (n-1)C(r-1)

2. Number of non-negative integer solutions for r integers adding up to n
= (n+r-1)C(r-1)
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Thank you Sir, had never really grasped these formulas until now.
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How many unique arrangements can you make with these characters?

**********||

There are 10 asterisks and two pipe characters. This is a method of solving this type of problem and it's known as partitioning.

The two pipe characters can be thought of as forming two barriers separating three groups. The stars in each group can be thought of as representing the the integer value in the first variable, second variable and third variable. For example, ****|****|** represents 4 + 4 + 2 = 10

So counting the combinations of the stars in three groups represents the possible unique equations that satisfy the conditions that add to 10.

\(\frac{12!}{(10!*2!)}\) = 66

Bunuel

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