Answer

Verified

447.3k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE