A combinatorial task I just can't solve Suppose you have $7$ apples, $3$ banana, $5$ lemons. How many options to form $3$ equal in size baskets ($5$ fruits in each) are exist?
At first I wrote: 
$\displaystyle \frac{15!}{7!3!5!} $
But its definitely not true because I do not care about permutations of apples in one $5$-sized basket. So the real value in divider is smaller than $7!$ (the same logic for other fruits).
Textbook says that the answer is $\displaystyle\frac{15!}{2^93^35^3}$
The method of getting of numerator is clear but divider...
 A: Assuming baskets are indistinguishable (as well as apples etc), there are surprisingly few arrangements possible. For apples the possible splits between baskets are: (5,2,0), (5,1,1), (4,3,0), (4,2,1), (3,3,1), (3,2,2).
For type (5,2,0) the possible splits of lemons are: (0,3,2), (0,2,3), (0,1,4), (0,0,5) (and in each case the bananas are distributed to make up the total of five fruits per basket).
For type (5,1,1) we have: (0,4,1), (0,3,2) (note that (0,1,4) would be duplicate).
For type (4,3,0) we have: (1,2,2), (1,1,3), (1,0,4), (0,2,3), (0,1,4), (0,0,5).
For type (4,2,1) we have: (1,3,1), (1,2,2), (1,1,3), (1,0,4), (0,3,2), (0,2,3), (0,1,4).
For type (3,3,1) we have: (2,2,1), (2,1,2), (2,0,3), (1,1,3), (1,0,4).
For type (3,2,2) we have: (2,3,0), (2,2,1), (1,3,1), (1,2,2), (0,3,2).
So a grand total of 29. 
A: First you have to choose 5 fruits for the first basket, you have $15\choose5$ combinations.
When you choose fruits for the second basket, now you have 10 fruits and you need to choose 5 again, so you have $10\choose5$ combinations.
Finally, you are left with 5 fruits and 1 basket, so there is $5\choose5$ which is 1 combination.
Now just multiply the results: $15\choose5$$\times$$10\choose5$$\times$$1$ which equals $756756$ and that's exactly the result from your textbook.
A: We need to be very careful about what we are assuming is distinguishable and what is not distinguishable. Both @almagest and @windircursed have correct answers with different sets of assumptions: 
@almagest assumed that both the baskets and the fruit are indistinguishable while 
@windircursed assumed that only the baskets were indistinguishable.
A third option would we to assume that all objects and baskets are distinguishable and then the correct answer would be simply $15!$.
Clearly after inspecting all the possibilities, we can see which interpretation the authors of the problem were assuming when they wrote their answer. Also the form of their answer was simply a way to mask how they came up with the answer: $2^93^35^3$ is simply the prime factorization of $(5!)^3$.
