Question

Find the number of solutions of the equation a+b+c+d=10 where a,b,c,d∈N

Find the number of solutions for the equation: x + y + z = 2002 where x, y, z ∈ N.

Find the number of solutions for the equation : 49m + 53n = 2020 where m, n ∈ N.

Find the number of solutions for the equation a + b + c = 289, where a, b, c are all 2-digit numbers.

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Answer 1

ANSWER

x+y+z=10

x,y,z⟶ positive

Number of Non negative for 

x,y,z⟶ positive 

⇒x1+x2+x3+............xr=n⇒n+r−1Cr−1x+y+z=10⟶for⇒a=x−1⇒b=y−1⇒c=z−1

a,b,c⟶non negative integer

⇒a+b+c=7⇒n=7⇒r=3

Therefore number of possible solutions are 

⇒7+3−1C3−1⇒9C2⇒29×8⇒36

Hence the correct answer is 36

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