Math logic

There are 20 children in the group. Every two children have a different name. Alena and John are among them. How many ways can we choose eight 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?

Final Answer:

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

Step-by-step explanation:

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

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

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

$d=2\cdot 17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11=196035840$

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