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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

Answer :

Answer:

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 = [tex]^{17}C_5 +( ^{17}C_4 \times ^{19}C_1} )+( ^{17}C_3 + ^{19}C_2})[/tex]

The ways of choosing = [tex]\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}[/tex]

[tex]=\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}[/tex]

[tex]=\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)!} )[/tex][tex]=6188 + ( 2380\times19 ) + (680 \times171)[/tex]

= 167688 ways

fichoh

Using the combination principle, the number of ways of selecting 5 students such that the number of boys is greater the number of girls selected ls 167688 ways.

  • Number of boys = 17

  • Number of girls = 19

Number of selections to be made = 5

In other to choose more boys than girls :

  • 5 boys and 0 girls
  • 4 boys and 1 girl
  • 3 boys and 2 girls

Using the combination relation :

(17C5 × 19C0) + (17C4 × 19C1) + (17C3 × 19C2)

(6188 × 1) + (2380 × 19) + (680 × 171)

(6188 + 45220 + 116280)

= 167688 ways

Therefore, the Number of ways of making the selection is 167688 ways

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