it because two variables are the same 1 and 1 so only change that can make the difference is 7 so 3C1 shows9 diff fruit pies
3 ppl ---- all gets odd no.
List of odd no. 1 3 5 7 _(9 not possible otherwise all will not get pies!)
Now , lets list possible ways the no. can be divided
1st per 2nd per 3rd per
1 3 5
1 5 3
1 7 1
7 1 1
1 1 7
3 1 5
3 3 3
(See that all combinations add to 9)
Now u c ms says:
1 1 7 = 9C1× 8C1× 7C7 (oe)× 3C1 =216 In above table:here c only 1 out of the three cn get 7 pies so 3C1.
1 3 5 = 9C1× 8C3× 5C5(oe)× 3! = 3024 In above table:here c either can hv 5 or 3 or 1 pie in any order so 3!
3 3 3 = 9C3× 6C3× 3C3 (oe) = 1680
In above table:here c its all getting 3 pie n order doesnt matter!
so you add all n get d final ans!
Hope i cud xplain! Though i myself is not much clear abt y we use 3C1 instd of 3! in the first part ...Bt dat was d explanation we were told @class . A little help there with this Alice123 or syed1995 ASTAR or any1!
this 1 1 7 or
171 or
711