<s>[INST] Combinatorics. Please explain me how to do it. In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively? This isn't a homework question. Please explain me the steps. Thank you!
 [/INST] If $A$ gives $a$ coins, clearly, $0\le a\le 6$
and $B+C=10-a$
Now, $0\le B\le 7\implies 0\le 10-a-C\le 7\implies 3-a\le C\le 10-a $ 
Also, $0\le C\le 8\implies$ max $(3-a,0)\le C\le $ min$(10-a,8)$
If $a=0,$   max $(3,0)\le C\le $ min$(10,8)\implies 3\le C\le 8$ so $C$ can assume $8-3+1=6$ values.
Similarly, for $a=1,2,3,4,5,6;$      $       C$ can assume $7,8,8,7,6,5$ values 
respectively.
So, the number possible combinations are $6+7+8+8+7+6+5=47$
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