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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

@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 ?

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....

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