Mathematics: A classroom of children has 17 boys and 19 girls in which five students are chosen to do presentations. In how many ways can the five students be chosen so that more boys than girls are selected

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A classroom of children has 17 boys and 19 girls

Ответ:

wickedskatboarder

06.02.2022

Mathematics

167688 ways

Step-by-step explanation:

Given that:

Number of boys = 17 &;

Number of girls = 19

The number of ways that five students can be chosen such that more boys than girls are selected are:

The ways of choosing = 5 boys 0 girls + 4 boys 1 girl + 3 boys 2 girls

The ways of choosing = $^{17}C_5 +( ^{17}C_4 \times ^{19}C_1} )+( ^{17}C_3 + ^{19}C_2})$

The ways of choosing = $\dfrac{17!}{5!(17-5)!} + \begin {pmatrix} \dfrac{17!}{4!(17-4)!} \times \dfrac{19!}{1!(19-1)!} \end {pmatrix} + \begin {pmatrix} \dfrac{17!}{3!(17-3)!} \times \dfrac{19!}{2!(19-2)!} \end {pmatrix}$

$=\dfrac{17!}{5!(12)!} + \begin {pmatrix} \dfrac{17!}{4!(13)!} \times \dfrac{19!}{1!(18)!} \end {pmatrix} + \begin {pmatrix} \dfrac{17!}{3!(14)!} \times \dfrac{19!}{2!(17)!} \end {pmatrix}$

$=\dfrac{17*16*15*14*13*12!}{5!(12)!} + ( \dfrac{17*16*15*14*13!}{4!(13)!} \times \dfrac{19*18!}{1!(18)!} ) + ( \dfrac{17*16*15*14!}{3!(14)!} \times \dfrac{19*18*17!}{2!(17)!} )$ $=6188 + ( 2380\times19 ) + (680 \times171)$

= 167688 ways

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A classroom of children has 17 boys and 19 girls in which five students are chosen to do presentations. In how many ways can the five students be chosen so that more boys than girls are selected

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