Use this, its basically the same thing
n+r-1 C r-1
N=5
R=3
So ans = 7C2
Now coming to your query.
How many people are there ? Three, who ? A B and C. Are these 3 people identical? Not possible !
How many donuts have ti be distributed ? 5 . Are these identical ? Yes . Heck, over counting will happen. Just keep this thought in mind and lets move forward.
For explaining to u in a better manner lets assume A B C be three bags and not people.
Further lets assume i am Donut 1 . Cool So mr soulja where you gonna jump ? I say, any one.
Same with other 4 Mr. Soulja's.
...
What this means is any donut can go in any bag , right ?
Bag A can have all 5 or even Zero.
So when such a distribution problem comes you can apply the illustrated method.
Now lets see what i exactly did there.
Lets introduce two sticks or separators or in our explanation i would like to call them partitions.
These partitions are a deciding factor.
How ?
A partition 1 B partition 2 C --- this combination reprsents any number of donuts in any bag. But at least one in each bag.
partition 1 partition 2 ABC ---- this combination means No donuts to anyone imagine this OOOOOABC
partition 1 A Partition2 BC --- this combination means at least one is before partition 1 , at least one is in A, and remaining can be distributed anywhere.
So basically this is a permutation of 7 items 2 idenctical partitions and 5 exact (donuts or Mr soulja's)
Hence, 7!/(5!2!)
Ill be happy to entertain further questions.
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If GMAT is Nasha , I am the BhamBhole