In how many ways we can divide 10 apples among 3 people in such a way that each should get at least 2 apples? i also want to know how to think of this type of problems if it was at least one apple or two or three what's the logic behind it 
 A: Assumption. Apples are identical but people are not.


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*Due to the assumption that apples are indistinguishable, we can give each person 2 apples so that we have $10-3\times2=4$ apples left. Therefore, the problem is reduced to counting the number of ways to distribute 4 apples among 3 people.

*The number of ways to distribute 4 apples among 3 people is $\binom{4+3-1}{3-1}=15.$ The argument used to establish this is often referred to as stars and bars, which says that there are $\binom{n+k-1}{k-1}$ ways to place $n$ identical stars into $k$ non-identical bins. In our case, the stars are the $n=4$ apples and the bins are the $k=3$ people.

A: 1)let's say the apples are identical 
W1 + 2 + w2 +2 + w3 +2 = 10 (w is whole no.) And wi represents no of apple ith person gets
W1 + w2 + w3 = 4 implying 6c2 ways(beggar's method)
2) they are not identical
10= 2+2+6 or 2+3+5 or 2+ 4 +4 or 3+3+4. 
We will make groups and arrange
10!/(2!2!6!2!)×3!(extra division of 2! Because there are 2 groups, If u know this then good but if u don't let me know I will explain)+10!(2!3!5!)*3!+10!/(2!4!4!2!)*3!+10!(3!3!4!2!)*3!
