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:
- 12
- 48
- 495
- 11880
- 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.
Best Answer
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).
Thus, the answer is
$$\frac{12\cdot 11 \cdot 10 \cdot 9}{4!} = 495$$