# From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition, including at least 4 boys and 4 girls. The 2 girls who won the prize last year should be included. In how many ways can the selection be made?

Last updated date: 25th Mar 2023

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Answer

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Hint: To solve the question, we have to find out the different cases of selection considering the given conditions and use combinations and factorial formula to calculate the number of ways of selecting 10 students for the competition for the different cases analysed.

Complete step-by-step answer:

The number of ways of selecting 10 students chosen for a competition from the 12 boys and 10 girls are selecting 5 boys and 5 girls or 6 boys and 4 girls or 4 boys and 6 girls.

Since the condition is given that the 10 students chosen for a competition, include at least 4 boys and 4 girls.

The other condition given is that the 10 students chosen for the competition must include the 2 girls who won the prize last year.

The number of ways of the 2 girls can be included in the 10 students chosen for a competition, are these 2 girls included in the group of 4 or 5 or 6 girls among the 10 students chosen for the competition.

The number of ways the selection can be made = Sum of ways of choosing 5 boys and 5 girls or 6 boys and 4 girls or 4 boys and 6 girls from the 12 boys and 10 girls of a class such that 2 girls are who won the prize last year are include in the team

\[={}^{12}{{C}_{5}}\times \left( {}^{8}{{C}_{3}}\times {}^{2}{{C}_{2}} \right)+{}^{12}{{C}_{6}}\times \left( {}^{8}{{C}_{2}}\times {}^{2}{{C}_{2}} \right)+{}^{12}{{C}_{4}}\times \left( {}^{8}{{C}_{4}}\times {}^{2}{{C}_{2}} \right)\]

\[={}^{12}{{C}_{5}}\times \left( {}^{8}{{C}_{3}}\times 1 \right)+{}^{12}{{C}_{6}}\times \left( {}^{8}{{C}_{2}}\times 1 \right)+{}^{12}{{C}_{4}}\times \left( {}^{8}{{C}_{4}}\times 1 \right)\]

We know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and \[n!=n\left( n-1 \right)\left( n-2 \right).....2\times 1\]

By applying the above formula, we get

\[=\dfrac{12!}{5!\left( 12-5 \right)!}\times \left( \dfrac{8!}{3!\left( 8-3 \right)!} \right)+\dfrac{12!}{6!\left( 12-6 \right)!}\times \left( \dfrac{8!}{2!\left( 8-2 \right)!} \right)+\dfrac{12!}{4!\left( 12-4 \right)!}\times \left( \dfrac{8!}{4!\left( 8-4 \right)!} \right)\]

\[=\dfrac{12!}{5!7!}\times \left( \dfrac{8!}{3!5!} \right)+\dfrac{12!}{6!6!}\times \left( \dfrac{8!}{2!6!} \right)+\dfrac{12!}{4!8!}\times \left( \dfrac{8!}{4!4!} \right)\]

\[\begin{align}

& =\dfrac{12\times 11\times 10\times 9\times 8\times 7!}{5!7!}\times \left( \dfrac{8\times 7\times 6\times 5!}{3!5!} \right)+\dfrac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6!6!}\times \left( \dfrac{8\times 7\times 6!}{2!6!} \right) \\

& +\dfrac{12\times 11\times 10\times 9\times 8!}{4!8!}\times \left( \dfrac{8\times 7\times 6\times 5\times 4!}{4!4!} \right) \\

\end{align}\]

Since we know that the above formula of \[n!\] can also be written as \[n!=n\left( n-1 \right)\left( n-2 \right)...(n-r)!\]

\[\begin{align}

& =\dfrac{12\times 11\times 10\times 9\times 8}{5!}\times \left( \dfrac{8\times 7\times 6}{3!} \right)+\dfrac{12\times 11\times 10\times 9\times 8\times 7}{6!}\times \left( \dfrac{8\times 7}{2!} \right) \\

& +\dfrac{12\times 11\times 10\times 9}{4!}\times \left( \dfrac{8\times 7\times 6\times 5}{4!} \right) \\

\end{align}\]

\[\begin{align}

& =\dfrac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2\times 1}\times \left( \dfrac{8\times 7\times 6}{3\times 2\times 1} \right)+\dfrac{12\times 11\times 10\times 9\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 1}\times \left( \dfrac{8\times 7}{2\times 1} \right) \\

& +\dfrac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}\times \left( \dfrac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1} \right) \\

\end{align}\]

By cancelling the common terms in denominator and numerator we get

\[=\left( 11\times 9\times 8 \right)\times \left( 8\times 7 \right)+\left( 11\times 3\times 4\times 7 \right)\times \left( 4\times 7 \right)+\left( 11\times 5\times 9 \right)\times \left( 5\times 2\times 7 \right)\]

\[=792\times 56+924\times 28+495\times 70\]

= 44352 + 25872 + 34650

= 104874

Thus, the number of ways of selecting 10 students chosen for the competition = 104874

Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.

Complete step-by-step answer:

The number of ways of selecting 10 students chosen for a competition from the 12 boys and 10 girls are selecting 5 boys and 5 girls or 6 boys and 4 girls or 4 boys and 6 girls.

Since the condition is given that the 10 students chosen for a competition, include at least 4 boys and 4 girls.

The other condition given is that the 10 students chosen for the competition must include the 2 girls who won the prize last year.

The number of ways of the 2 girls can be included in the 10 students chosen for a competition, are these 2 girls included in the group of 4 or 5 or 6 girls among the 10 students chosen for the competition.

The number of ways the selection can be made = Sum of ways of choosing 5 boys and 5 girls or 6 boys and 4 girls or 4 boys and 6 girls from the 12 boys and 10 girls of a class such that 2 girls are who won the prize last year are include in the team

\[={}^{12}{{C}_{5}}\times \left( {}^{8}{{C}_{3}}\times {}^{2}{{C}_{2}} \right)+{}^{12}{{C}_{6}}\times \left( {}^{8}{{C}_{2}}\times {}^{2}{{C}_{2}} \right)+{}^{12}{{C}_{4}}\times \left( {}^{8}{{C}_{4}}\times {}^{2}{{C}_{2}} \right)\]

\[={}^{12}{{C}_{5}}\times \left( {}^{8}{{C}_{3}}\times 1 \right)+{}^{12}{{C}_{6}}\times \left( {}^{8}{{C}_{2}}\times 1 \right)+{}^{12}{{C}_{4}}\times \left( {}^{8}{{C}_{4}}\times 1 \right)\]

We know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and \[n!=n\left( n-1 \right)\left( n-2 \right).....2\times 1\]

By applying the above formula, we get

\[=\dfrac{12!}{5!\left( 12-5 \right)!}\times \left( \dfrac{8!}{3!\left( 8-3 \right)!} \right)+\dfrac{12!}{6!\left( 12-6 \right)!}\times \left( \dfrac{8!}{2!\left( 8-2 \right)!} \right)+\dfrac{12!}{4!\left( 12-4 \right)!}\times \left( \dfrac{8!}{4!\left( 8-4 \right)!} \right)\]

\[=\dfrac{12!}{5!7!}\times \left( \dfrac{8!}{3!5!} \right)+\dfrac{12!}{6!6!}\times \left( \dfrac{8!}{2!6!} \right)+\dfrac{12!}{4!8!}\times \left( \dfrac{8!}{4!4!} \right)\]

\[\begin{align}

& =\dfrac{12\times 11\times 10\times 9\times 8\times 7!}{5!7!}\times \left( \dfrac{8\times 7\times 6\times 5!}{3!5!} \right)+\dfrac{12\times 11\times 10\times 9\times 8\times 7\times 6!}{6!6!}\times \left( \dfrac{8\times 7\times 6!}{2!6!} \right) \\

& +\dfrac{12\times 11\times 10\times 9\times 8!}{4!8!}\times \left( \dfrac{8\times 7\times 6\times 5\times 4!}{4!4!} \right) \\

\end{align}\]

Since we know that the above formula of \[n!\] can also be written as \[n!=n\left( n-1 \right)\left( n-2 \right)...(n-r)!\]

\[\begin{align}

& =\dfrac{12\times 11\times 10\times 9\times 8}{5!}\times \left( \dfrac{8\times 7\times 6}{3!} \right)+\dfrac{12\times 11\times 10\times 9\times 8\times 7}{6!}\times \left( \dfrac{8\times 7}{2!} \right) \\

& +\dfrac{12\times 11\times 10\times 9}{4!}\times \left( \dfrac{8\times 7\times 6\times 5}{4!} \right) \\

\end{align}\]

\[\begin{align}

& =\dfrac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2\times 1}\times \left( \dfrac{8\times 7\times 6}{3\times 2\times 1} \right)+\dfrac{12\times 11\times 10\times 9\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 1}\times \left( \dfrac{8\times 7}{2\times 1} \right) \\

& +\dfrac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}\times \left( \dfrac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1} \right) \\

\end{align}\]

By cancelling the common terms in denominator and numerator we get

\[=\left( 11\times 9\times 8 \right)\times \left( 8\times 7 \right)+\left( 11\times 3\times 4\times 7 \right)\times \left( 4\times 7 \right)+\left( 11\times 5\times 9 \right)\times \left( 5\times 2\times 7 \right)\]

\[=792\times 56+924\times 28+495\times 70\]

= 44352 + 25872 + 34650

= 104874

Thus, the number of ways of selecting 10 students chosen for the competition = 104874

Note: The possibility of mistake can be the calculation mistake since the procedure of solving involves large calculations. Hence be very careful in simplifying and cancelling the terms to get the desired result.

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