How to distinguish between combination and permutation questions?

Solution 1:

Example: How many different committees of 4 students can be chosen from a group of 15? Answer: There are possible combinations of 4 students from a set of 15.

In your first example you are choosing slots for each student to occupy, but you don't care which order the students make in each slot. You only care about the combination.

Example: How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15P4 possible permutations of 4 students from a group of 15.

In permutations the order of the objects matters. If the objects are indistinguishable then it is a combinations problem.

In your second example you can distinguish one student from another; therefore it is permutations problem, so switching the order of the students in the line will affect the way the photograph looks.

To tell the difference you look at the 'object' in question and ask yourself is it distinguishable or not in the context of the question.

Solution 2:

${}_nC_r$ is the number of ways of selecting $r$ items from a set of $n$ items.

${}_nP_r$ is the number of ways of selecting $r$ items from a set of $n$ items and arranging the $r$ selected items in all possible orders.

In brief, combinations are used when you are selecting only; permutations are used when you are selecting and arranging.

It is often said that combinations are used when order doesn't matter and that permutations are used when order matters, but this can be misleading. By "order matters", people probably mean that different orders are distinguishable, which is important, but not the only issue. The other important issue is whether all possible orders are allowed.

If, in selecting $r$ items from $n$, different orders are distinguishable and all possible orders are allowed, then ${}_nP_r$ is appropriate. If, in selecting $r$ items from $n$, either different orders are not distinguishable or only only one order is allowed, then ${}_nC_r$ would be appropriate.

Here are some examples.

1. How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: to find all possible lineups, you have to find all possible selections, and for each selection, you have to find all possible ways of lining up the selected students. Different lineups are distinguishable and all possible lineups can occur. So there are ${}_{15}P_4$ lineups.
2. How many ways can 4 students from a group of 15 be lined up for a photograph in order by age, with oldest on the left and youngest on the right? Answer: to find all possible lineups, you have to find all possible selections. Different lineups are distinguishable, but only one lineup is allowed. Once you determine the selection selection, the lineup is forced, since it is according to age. Hence there are ${}_{15}C_4$ lineups.
3. How many different committees of four students can be chosen from a group of 15? Answer: committees are defined only by their membership, so different orders of the selected committee members are not distinguishable. Hence there are ${}_{15}C_4$ committees.
4. There are three E Scrabble tiles and four S Scrabble tiles. In how many ways can all of the tiles be placed side-by-side along a straight line? Answer: physical pieces of wood are distinguishable, and so different placements of them are distinguishable. Futhermore all possible placements are possible. Hence there are ${}_7P_7=7!$ placements.
5. How many different "words" can be spelled using the letter E three times and the letter S four times, where any sequence of letters counts as a word? Answer: a word is distinguished by the sequence of letters, that is, the order, but is not tied to the placement of any physical artifacts. Different sequences of Es and Ss are distinguishable, but the result of permuting only the Es or only the Ss in a particular sequence does not give a distinguishable result. So different orders within the subsequence of Es or within the subsequence of Ss are not distinguishable. Hence there are $$\frac{{}_7P_7}{{}_3P_3\,{}_4P_4}=\frac{7!}{3!\,4!}$$ words. This answer equals ${}_7C_4$, which can be understood as follows: to specify a word, you need only state which four letter positions among the 7 contain Ss, with the understanding that every other position contains an E. A set of four letter positions is distinguished only by its elements, and not by the order in which they are listed. So having Ss in positions 2, 3, 5, and 7 is indistinguishable from having Ss in positions 5, 3, 7, and 2. In other words, different orders of the four S positions are not distinguishable. Hence there are ${}_7C_4$ words.

Solution 3:

In general whenever the position matters, the question is a permutation question. For example when (A,B) and (B,A) are different, it is a permutation question.

And whenever the position does not matter, it is a combination question. For example (A,B) is the same as (B,A).

Example :

1) In how many ways can we select 2 students from three students A,B,C?

Answer = $^3C_2 = \frac{3!}{2!\times(3-2)!} = 3$

This is a combination question because we can select in three ways as follows : (A,B) , (B,C) , (C,A). See here, when (A,B) is selected , it is the same as (B,A).

2) In how many ways can 2 students be lined in a row, out of a total of 3 students A,B,C ?

Answer = $^3 P_2 = \frac{3!}{(3-2)!} = 6$

This is a permutation question because there are six ways as follows: (A,B) , (B,A) , (B,C) , (C,B) , (A,C) , (C,A). Note that now (A,B) and (B,A) are different as in (A,B) , A gets the first position in the row and B gets the second position in the row. While in (B,A) , B gets the first position in the row and A gets the second position. Hence we have to count them as separate possibilities, unlike in the previous example.