In how many ways can four students be chosen from a group of 12 students?

Myself and my Math teacher are at a disagreement in to what the proper method of solving the question In how many ways can four students be chosen from a group of 12 students? is.

The question comes straight out of a Math revision sheet from a Math book distributed under the national curriculum. The options it gives for answers are:

1. 12
2. 48
3. 495
4. 11880
5. 40320

As we are currently learning Permutations and Combinations, my above interpretation is that it is asking for a Combination without repetition or $\frac{(n+r-1)!}{r!(n-1)!}$ which gives you the amount of combinations without repetition (as you cannot pick the same student twice.) Now my teacher argues that the answer the book provides is correct. The books answer simply says to use $^{n}C_{r}$ or $\frac{n!}{r!(n-r)!}$.

What is the correct method of answering this? The book states 3. 495 is the answer.

• 495 is the correct answer. I think that you might be mixing up the formulas for combination with repetition and combination without repetition.
– GAM
Oct 30 '13 at 23:40

You have 12 choices for the first student chosen, 11 choices for the next, then 10, then 9. However, this over-counts everything by a factor of 4! (the number of ways in which four objects can be arranged with regard to order).

$$\frac{12\cdot 11 \cdot 10 \cdot 9}{4!} = 495$$

• Does the "without repetition" refer to repetition within each combination/permutation? Or does it refer to repetition over all? Oct 30 '13 at 23:40
• The phrase 'without repetition' usually means either that you can't pick the same person twice, or that you can't count the same ordering twice. For example, once you have counted the group of students ABCD, you shouldn't also count BCDA: that would involve the repetition of re-counting the same group. To avoid repetition altogether, you need to make sure you count: ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, ... etc only once. Hence the division by 4!. Oct 30 '13 at 23:45
• @jduncanator : It means that if you are picking "students" who are named $a_1,a_2, a_3, \dots, a_{12}$, then $\{a_1,a_4,a_5,a_7\}$ is a legitimate pick, but $\{a_1,a_1,a_2,a_3\}$ is not. Because if you pick $a_1$ twice, then you are really only choosing three students. Oct 30 '13 at 23:45
• @BenjaminDickman Thanks, that makes sense! I was wondering where that division by 4! came from! Oct 30 '13 at 23:59
• @PatrickDaSilva That was what I thought "without repetition" meant... however in this scenario, it seems not to be the case. Oct 31 '13 at 0:00

To justify the $^nC_r$ formula, you can line up all the $n$ boys in $n!$ ways. Then you take the first $r$ for your selection. You can scramble the first $r$ and the last $n-r$ in any way and get the same selection, so you divide by $r!$ and by $(n-r)!$ to get the formula.

As asked, the question is seeking to know the number of subsets of size 4 in a set of size 12. A subsequent accounting for the order in which they are chosen is a change in the nature of the question. So there are $${12\choose 4}$$ ways to do this. Your interpretation is correct.

Were order important, you would be asked the number of ordered subsets of size four, or you might be asked to name them something like president, vice-president, secretary and treasurer. This labeling can be done to each subset of size 4 freely, so it boosts the count by a factor of $4! = 24.$ No such secondary labeling is present in the question.

Being asked to choose "a group (set) of size 4" implies sampling without replacement.

In a word, the book is on rock-solid ground.