Answer:
80730
Step-by-step explanation:
The answer is c(27,5)=27*26*25*24*23/5!. Here is why.
Let us solve a simpler problem. In how many ways 25 identical balls can be placed in six distinct urns such that no urn is empty?
25 balls in a row make 24 places for five dividers, so the answer is c(24,5).
2. Next problem. In how many ways 50 identical balls can be placed in six distinct urns such that no urn is empty and each urn contains even number of balls.
We group 50 balls in 25 pairs, so the answer is the same as in the previous problem which is
c(48/2,5)=c(24,5).
Let us denote the content of each urn as
u(1),u(2),…u(6). Note that each u(i) is even. Denote u(i) as u(i)=2k(i) where k is a natural number. We have
2k(1)+2k(2)+2k(3)+2k(4)+2k(5)+2k(6)=50.
3. Now let us proceed to the original problem where odd number of balls in each urn is required.
We have
2k(1)-1+2k(2)-1+2k(3)-1+2k(4)-1+2k(5)-1+2k(6)-1=50 or
2k(1)+2k(2)+2k(3)+2k(4)+2k(5)+2k(6)=56.
So the original problem reduces to previous problem where even number of balls in each urn is required but the total number of balls is 56 instead of 50.
Its answer is c(54/2,5)=c(27,5).
