26. How many ways are there to put 7 identical apples into 4 identical packages so that each package has at least one apple?
26. Answer: 350 .
By considering the numbers of apples in the packages, there are 3 cases:
1) $(4,1,1,1)$. ( $\left.\begin{array}{l}7 \\ 4\end{array}\right)=\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}=35$.
2) $(3,2,1,1)$. $\binom{7}{3}\binom{4}{2}=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=\times \frac{4 \times 3}{2 \times 1}=35 \times 6=210$.
3) $(2,2,2,1)$. $\frac{1}{3!}\binom{7}{2}\binom{5}{2}\binom{3}{2}=\frac{1}{6} \times \frac{7 \times 6}{2 \times 1} \times \frac{5 \times 4}{2 \times 1} \times \frac{3 \times 2}{2 \times 1}=105$.
So the total number of ways is $35+210+105=350$.