Query in P&C

for sum = 9, it worked cleanly because after giving 1 to each die, you got

a + b + c + d = 5 and the upper bound (≤5) didn’t matter

but when the sum becomes bigger like 12, the upper bound starts mattering

let’s see how to think about it

sum = 12

give 1 to each → let xi = ai − 1

so:

x1 + x2 + x3 + x4 = 12 − 4 = 8

and now each xi can be from 0 to 5 (because original die max is 6)

step 1 (ignore upper bound):

number of non-negative solutions = C(8+4−1, 3) = C(11,3)

step 2 (remove invalid cases where any xi > 5)

if any xi ≥ 6, say x1 ≥ 6

put x1 = y1 + 6

then:

y1 + x2 + x3 + x4 = 2

number of such solutions = C(2+4−1,3) = C(5,3)

this can happen for any of the 4 variables → multiply by 4

step 3 (check double violations)

can two variables be ≥6?

no, because 6+6 = 12 > 8, impossible

so no need to add back anything

final answer:

C(11,3) − 4×C(5,3)

= 165 − 4×10

= 165 − 40 = 125

so the key idea is:

small sums → upper bound doesn’t matter

bigger sums → use stars & bars minus cases where variable exceeds limit

Watch the afterboards topic video on PnC, titled under "No of solutions", you can learn the method from there.

Else the general form for such equations is, (n+r-1)C(r-1)

(Stars and bars method or as Bhavesh Bhaiya calls it pp method^^)