Seating pupils

We will work with a class in which there are 30 pupils, 40% of them are boys, the number of benches is 18. Determine the number of possibilities in the following tasks.

1) Determine in how many ways it is possible to select for a competition a trio of pupils, if it is not specified how many boys and how many girls.

2) Determine in how many ways it is possible to seat the pupils in the class if it is all the same to all of them where they sit.

3) Determine in how many ways the pupils can be seated, if always only girls or boys sit together.

4) Determine in how many ways the class self-government can be chosen, consisting of the chairperson, vice-chairperson, and treasurer.

Final Answer:

n₁ =  4060
n₂ =  105
n₃ =  219
n₄ =  24360

Step-by-step explanation:

$$\begin{gathered} C_3(30)=\left(\genfrac{}{}{0ex}{}{30}{3}\right)=\frac{30!}{3!(30-3)!}=\frac{30\cdot 29\cdot 28}{3\cdot 2\cdot 1}=4060 \\ n=30 \\ q=40\%=\frac{40}{100}=\frac{2}{5}=0.4 \\ c=q\cdot n=0.4\cdot 30=12 \\ d=n-c=30-12=18 \\ l=18 \\ {n}_1=\left(\genfrac{}{}{0ex}{}{n}{3}\right)=4060 \end{gathered}$$

$$\begin{gathered} C_2(15)=\left(\genfrac{}{}{0ex}{}{15}{2}\right)=\frac{15!}{2!(15-2)!}=\frac{15\cdot 14}{2\cdot 1}=105 \\ l>n/2 \\ {n}_2=\left(\genfrac{}{}{0ex}{}{n/2}{2}\right)=\left(\genfrac{}{}{0ex}{}{30/2}{2}\right)=105 \end{gathered}$$

$$\begin{gathered} C_2(18)=\left(\genfrac{}{}{0ex}{}{18}{2}\right)=\frac{18!}{2!(18-2)!}=\frac{18\cdot 17}{2\cdot 1}=153 \\ {C}_2(12)=\left(\genfrac{}{}{0ex}{}{12}{2}\right)=\frac{12!}{2!(12-2)!}=\frac{12\cdot 11}{2\cdot 1}=66 \\ {n}_3=\left(\genfrac{}{}{0ex}{}{d}{2}\right)+\left(\genfrac{}{}{0ex}{}{c}{2}\right)=153+66=219 \end{gathered}$$

$n_4=n\cdot (n-1)\cdot (n-2)=30\cdot (30-1)\cdot (30-2)=24360$

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