My dear ladybird ldd
Let's analyse your inference:
4 oranges are to be distributed to 4 people - If I look from oranges' perspective then 1st orange have 4 choices (as it can be distributed to anyone of the four people), now what about 2nd orange? Does it have lesser choices - certainly not. It also has 4 choices and that's also true for remaining two oranges.
Now appears one more doubt that should not answer be 4^{4} then? Look, in our counting we have assumed that all the oranges are different (that's why we are able to name them 1st, 2nd, .. and so on) but actually it's not. So in above counting we have a lot of repetitions which need to be eradicated. And that's a tough task.
So think alternately. We are having 4 identical oranges which are to be distributed to four (certainly) different people. That is, we are to just divide the 4 oranges in four parts and distribute them to four people (say 1st part belongs to 1st person, 2nd part belongs to 2nd person,.. and so on) Now to divide four oranges in four parts, we are to just place 3 partitions anywhere between those 4 oranges. Or we need to select 3 (distinct) places where the partitions will be placed out of total 7 places available (4 for oranges and 3 for partitions).
And this can be easily achieved in C(7, 4) =^{ 7}C_{4 }= 35 ways.
Hope it is clear.
Kamal Lohia^{ } |