# 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?

Result

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

#### Solution:

$p_{ 1 } = 7! = 5040 \ \\ a = 1 \cdot \ 19 \cdot \ 18 \cdot \ 17 \cdot \ 16 \cdot \ 15 \cdot \ 14 \cdot \ 13/p_{ 1 } = 1 \cdot \ 19 \cdot \ 18 \cdot \ 17 \cdot \ 16 \cdot \ 15 \cdot \ 14 \cdot \ 13/5040 = 50388$
$p_{ 2 } = 6 \cdot \ 5 \cdot \ 4 \cdot \ 3 \cdot \ 2 \cdot \ 1 = 720 \ \\ b = 1 \cdot \ 1 \cdot \ 18 \cdot \ 17 \cdot \ 16 \cdot \ 15 \cdot \ 14 \cdot \ 13/p_{ 2 } = 1 \cdot \ 1 \cdot \ 18 \cdot \ 17 \cdot \ 16 \cdot \ 15 \cdot \ 14 \cdot \ 13/720 = 18564$
$p_{ 3 } = 7 \cdot \ 6 \cdot \ 5 \cdot \ 4 \cdot \ 3 \cdot \ 2 \cdot \ 1 \cdot \ 2 = 10080 \ \\ c = 3 \cdot \ 19 \cdot \ 18 \cdot \ 17 \cdot \ 16 \cdot \ 15 \cdot \ 14 \cdot \ 13/p_{ 3 } = 3 \cdot \ 19 \cdot \ 18 \cdot \ 17 \cdot \ 16 \cdot \ 15 \cdot \ 14 \cdot \ 13/10080 = 75582$
$d = 2 \cdot \ 17 \cdot \ 16 \cdot \ 15 \cdot \ 14 \cdot \ 13 \cdot \ 12 \cdot \ 11 = 196035840 = 1.960358\cdot 10^{ 8 }$

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