#non-negative Integer solutions

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21)

3 cases possible

case i) x1 > 40

so smallest value of x1 is 41

x1 + 41 + x2 +  x3 = 100

x1 + x2 + x3 = 59

so possible combinations = ^59+3-1C₅₉ = ⁶¹C₅₉ = 1830

case ii) x2 > 40 and x1<=40

first we will find number of combination such that x2 > 40 and then subtract the combinations of x2 > 40 and x1 > 40

first we will find number of combination such that x2 > 40

x1 + x2 + 41 + x3 = 100

x1 + x2 + x3 = 59

so possible combinations = ^59+3-1C₅₉ = ⁶¹C₅₉ = 1830

Now number of combination such that x2 > 40 and x1 > 40

x1 + 41 + x2 + 41 + x3 = 100

x1 + x2 + x3 = 18

so possible combinations = ^18+3-1C₁₈ = ²⁰C₁₈ = 190

Hence number of combination such that x2 > 40 and x1<=40 = 1830 - 190 = 1640

case iii) x3 > 40 and x1 <= 40 and x2 <= 40

first we will find number of combination such that x3 > 40 and then subtract the combinations of x3 > 40 and x1 > 40 and also subtract the combinations of x3 > 40 and x2 > 40

first we will find number of combination such that x3 > 40

x1 + x2 + x3 + 41 = 100

x1 + x2 + x3 = 59

so possible combinations = ^59+3-1C₅₉ = ⁶¹C₅₉ = 1830

Now number of combination such that x3 > 40 and x1 > 40

x1 + 41 + x2 + x3 + 41= 100

x1 + x2 + x3 = 18

so possible combinations = ^18+3-1C₁₈ = ²⁰C₁₈ = 190

Now number of combination such that x3 > 40 and x2 > 40

x1 + x2 + 41 + x3 + 41= 100

x1 + x2 + x3 = 18

so possible combinations = ^18+3-1C₁₈ = ²⁰C₁₈ = 190

Hence number of combination such that  x3 > 40 and x1 <= 40 and x2 <= 40 = 1830 - 190 - 190 = 1450

So total ways = 1830 + 1640 + 1450 = 4920

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@Gabbar see this answer to 20th question

x1 + x2 + x3 + x4 =100

now x1 can take value from 0 to 10. So we will substitute the values one by one and find number of combinations

1) 0 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 100 => ^100+3-1C₁₀₀ = ¹⁰²C₁₀₀ = $\frac{101*102}{2}$

2) 1 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 99 => ^99+3-1C₉₉ = ¹⁰¹C₉₉ = $\frac{100*101}{2}$

3) 2 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 98 => ^98+3-1C₉₈ = ¹⁰⁰C₉₈ = $\frac{99*100}{2}$

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and so on

11) 10 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 90 => ^90+3-1C₉₀ = ⁹²C₉₀ = $\frac{91*92}{2}$

now we will have to add all these

total combination = summation of  n(n+1)/2 where n from 91 to 101

                          = summation of $\frac{n^{2}+n}{2}$

                          = $\frac{n(n+1)(2n+1)}{6*2}$ + $\frac{n(n+1)}{2*2}$

first put n=101 and then subtract by putting n=91 we get

total combination = 176851 - 129766 = 47085

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2nd method)

x1 + x2 + x3 + x4 =100

total combination = ^100+4-1C₁₀₀ = ¹⁰³C₁₀₀ = 176851

now to get the desired result subtract the case that value of x1 is atleast 11

11 + x1 + x2 + x3 + x4 = 100

x1 + x2 + x3 + x4 = 89

total combination = ^89+4-1C₈₉ = ⁹²C₈₉ = 125580

So ans = 176851 - 125580 = 51271

So which one is correct and why ?

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I have used second one.

47085 when u take atleast 10 instead of 11 that was my silly mistake when I have solved it first time....