Q1. How many ways are there to distribute 4 distinct oranges and 6 distinct apples into 5 distinct boxes?
Approach 1: Let us assume the two tasks disjoint. So oranges go in $5^4$ ways and apples in $5^6$ ways. So total ways in which both tasks happen together is $5^4*5^6=5^{10}$
Approach 2: Assume there are together 10 distinct objects to be distributed. Answer is $5^{10}$
Q2. How many ways are there to distribute 4 identical oranges and 6 distinct apples into 5 distinct boxes?
Approach 1: The two tasks are disjoint. orange task can be done in $\binom{4+5-1}{4}$ Using Sticks and Stones for Counting number of Ways. apple task is as before. so total number of ways $=\binom{4+5-1}{4}*5^6$
Approach 2: But if i combine both tasks as in Q1, first assume that oranges are different then i have total number of ways is $5^4*5^6=5^{10}$. Now since oranges are identical, so total number of ways should be $5^{10}/4!$. This is not even a whole number.
So $\binom{4+5-1}{4}*5^6\neq 5^{10}/4!$
Why is approach 2 going wrong in Q2 and not Q1?
Best Answer
It is not always true that $4!$ ways of putting distinct oranges in correspond to one way of putting identical oranges in. This is only true if one orange was put into each of four boxes, but all four oranges may be put into one box, or two in one and two in another, and so on. The relevant divisor is not uniform across all cases.