Students, from a class of $30$, with an equal number of girls and boys, are electing the members of their class council. They are electing four representatives from their class: a president, a vice president, a treasurer, and a florist. Each student can hold only one position, and the students have agreed that the council should consist of an equal number of girls and boys. Tom has calculated how many variations of the council's composition are possible.
Tom’s solution:
(1) In the class, there are $15$ girls and $15$ boys, and $4$ council members are to be selected from them. Therefore, the council will consist of $2$ girls and $2$ boys.
(2) The number of possible selections of $2$ girls from $15$ can be determined as the number of ordered pairs from $15$ elements, which is $15\cdot 14 = 210$. Similarly, we can determine the number of possible selections of $2$ boys from $15$, which is also $15\cdot 14=210$.
(3) Thus, there are $210$ ways to select a pair of girls, and $210$ ways to select a pair of boys. In total, we get $210\cdot 210=44\,100$ different sets of $4$ council members.
(4) Now, we need to consider that for each set of $4$ council members, there are $4!$ ways to assign them to the different council positions. The total number of possible variations of the class council's compositions is $44\,100\cdot 4!=1\,058\,400$.
Tom's solution is incorrect. Determine what the correct solution should look like and indicate where the mistake is made.
The mistake is in step (2). With the chosen method of selecting a pair of girls and a pair of boys, their order does not matter. Tom should have calculated their number as the number of unordered pairs from $15$ elements ($2$-combinations of $15$ element set), i.e. ${15 \choose 2}=105$. The number of possible sets of four members is then $105\cdot 105=11\,025$. The final number of variants of the class council (according to step 4) is $11\,025\cdot 4!=264\,600$.
The mistake is in step (4). For each set of four council members, there are $2!\cdot 2!$ ("number of permutations of girls" ∙ "number of permutations of boys") ways to assign them roles. The total number of variants of the class council is $44\,100 \cdot 2!\cdot 2!=167\,400.$
The mistake is in step (3). There are $210$ ways to select a pair of girls and 210 ways to select a pair of boys. Thus, we have a total of $210+201=420$ different sets of four council members. The final number of variants of the class council (according to step 4) is then $420\cdot 2!=10\,800$.
The mistake is in step (2). Tom should have calculated the number of pairs of girls and boys as the number of ordered pairs with repetition from $15$ elements ($2$-permutations with repetition from $15$ element set), i.e. $15^2=225$. The number of possible sets of four members is then $225\cdot 225=50\,625$. The final number of variants of the class council (according to step 4) is then $50\,625\cdot 4!=1\,215\,000$.