Given, 30⋅‌30C0+29⋅‌30C1+...+2⋅‌30C28+‌30C29=n⋅2m This can be written as,
29
∑
r=0
(30−r)30Cr=n⋅2m or ‌‌
30
∑
t=0
(30−r)⋅‌30Cr=n⋅2m ⇒‌‌
30
∑
r=0
30⋅‌30Cr−
30
∑
r=0
r⋅‌30Cr=n⋅2m ⇒‌‌30‌
30
∑
r=0
‌30Cr−
30
∑
r=0
r⋅‌30Cr=n⋅2m Using combination properties, ‌30⋅(2)30−30⋅(2)29‌=n⋅2m ⇒‌30⋅(2)29(2−1)‌=n⋅2m ⇒‌2.15⋅(2)29‌=n⋅2m ⇒‌15⋅(2)30‌=n⋅2m Comparing both sides, n=15‌ and ‌m=30 ⇒‌‌n+m=15+30=45