Question

There are \[5\] girls and \[7\] boys. A team of \[3\] boys and \[2\] girls is to be formed such that no two specific boys are in the same team. What is the number of ways to do so?
A. \[400\]
B. \[250\]
C. \[200\]
D. \[300\]

Answer 1

Hint:Find the number of ways for selecting \[3\] boys and \[2\] girls from \[5\] girls and \[7\] boys. Multiply these two numbers of ways for getting all possible numbers of teams can be formed. If two specific boys are in the same team, then you have to select another boy from \[\left( {7 - 2} \right) = 5\] boys and the number of ways for selecting the girls remains the same. Find the number of ways for selecting a boy from \[5\] boys and multiply it by the number of ways for selecting girls. Subtract it from the total number of teams to get the desired answer.

Formula used
Number of ways for selecting \[r\] number of objects from \[n\] number of objects is \[{}^n{C_r}\]
\[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
If there are \[a\] number of ways for selecting an object and \[b\] number of ways for selecting another object, then the total number of ways for selecting the objects is \[ab\].

Complete step by step solution
There are \[5\] girls and \[7\] boys in the group.
A team of \[3\] boys and \[2\] girls is to be formed.
Since \[3\] boys are to be selected from \[7\] boys, so total number of ways for doing that is \[{}^7{C_3} = \dfrac{{7!}}{{\left( {7 - 3} \right)!3!}} = \dfrac{{7!}}{{4!3!}} = \dfrac{{7 \times 6 \times 5 \times 4!}}{{4! \times 3 \times 2 \times 1}} = \dfrac{{210}}{6} = 35\]
Since \[2\] girls are to be selected from \[5\] girls, so the total number of ways for doing that is
\[{}^5{C_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!2!}} = \dfrac{{5!}}{{3!2!}} = \dfrac{{5 \times 4 \times 3!}}{{3! \times 2 \times 1}} = \dfrac{{20}}{2} = 10\]
\[\therefore \]Total number of possible teams can be formed is \[35 \times 10 = 350\]
Now, if two specific boys are selected in the same team, then another boy is to be selected from the group of \[\left( {7 - 2} \right) = 5\] boys.
Number of ways for doing that is \[{}^5{C_1} = \dfrac{{5!}}{{\left( {5 - 1} \right)!1!}} = \dfrac{{5!}}{{4!1!}} = \dfrac{{5 \times 4!}}{{4! \times 1}} = 5\]
The number of ways for selecting girls remains the same.
\[\therefore \]Total number of such teams can be formed is \[5 \times 10 = 50\]
\[\therefore \] Required number of ways is \[350 - 50 = 300\]
Hence option D is correct.

Note: Number of ways for selecting \[r\] number of objects from \[n\] number of objects is \[{}^n{C_r}\] but number of ways for selecting and arranging \[r\] number of objects from \[n\] number of objects is \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]. If there are some number of ways for selecting different objects and you have to select all the different objects then you have to multiply all the number of ways but if you have to select any one or some particular objects then you have to add the number of ways.