Math logic

There are 20 children in the group, each two children have a different name. Alena and John are among them. How many ways can we choose 8 children to be among the selected

A) was John
B) was John and Alena
C) at least one was Alena, John
D) maximum one was Alena, John?

Correct result:

a =  50388
b =  18564
c =  75582
d =  196035840

Solution:

$$\begin{gathered} p_1=7!=5040 \\ a=1\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\mathrm{/}{p}_1=1\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\mathrm{/}5040=50388 \end{gathered}$$

$$\begin{gathered} p_2=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720 \\ b=1\cdot 1\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\mathrm{/}{p}_2=1\cdot 1\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\mathrm{/}720=18564 \end{gathered}$$

$$\begin{gathered} p_3=7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2=10080 \\ c=3\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\mathrm{/}{p}_3=3\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\mathrm{/}10080=75582 \end{gathered}$$

$d=2\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11=196035840=1.960358\cdot 10^8$

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