selecting students for mathematics and language competitions

Question

In a class of $25$ students, there are $13$ boys and $12$ girls. From this class, we randomly choose a team of $5$ students to represent the class in a mathematics competition.

(a) Find the different ways can you choose $5$ students to represent the class.

My answer is $6375600$.

(b) Find the probability that the team consists $3$ girls and $2$ boys.

My answer is $32.3\%$.

From the same class, we are going to randomly choose a team of $7$ students to represent in a language competition. It is possible for all students to participate in both competitions.

(c) Calculate the probability that exactly $3$ students are chosen to compete in both competitions.

Answer 1

• HINT: When you select a team of 5 people, it does not matter which order you choose, so several times your answer is taken into account.

P.S. read about Combinations

Answer 2

Your (a) is wrong. Although you didn't provide how you got the answer, but I guess you did

$$25\cdot24\cdot23\cdot22\cdot21,$$

which is permutation of five out of twenty five students in a row, but the correct step is

$${25\choose5},$$

and notice that this kind of error can be identified by adapting combinatorics symbol as above, since

$${25\choose5}\cdot5!=6375600.$$

(b) I recommend write down your thinking process first, so

$$(\textrm{choose three girls})\cdot(\textrm{choose two boys})\\ ={12\choose3}\cdot{13\choose2},$$

so the probability is

$$\frac{\textrm{answer of question (b)}}{\cdot\ \textrm{of (a)}}=\Large\frac{{12\choose3}\cdot{13\choose2}}{25\choose5}\approx32.3\%,$$

and mysteriously you're correct.

Finally consider (c),

$$\frac{(\textrm{choose three lucky guys})\cdot(\textrm{others for math})\cdot(\textrm{others for lang.})}{(\textrm{choose math})\cdot(\textrm{choose lang.})}$$ $$=\frac{{25\choose3}\cdot{22\choose2}\cdot{20\choose4}}{{25\choose5}\cdot{25\choose7}}.$$