The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is

If for some $$m, n ;{ }^6 C_m+2\left({ }^6 C_{m+1}\right)+{ }^6 C_{m+2}>{ }^8 C_3$$ and $${ }^{n-1} P_3:{ }^n P_4=1: 8$$, then $${ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m$$ is equal to

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

Let $$\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$$ and $$\beta=\frac{(5 !) !}{(5 !)^{4 !}}$$. Then :