# How can we tell from looking at a problem that multiplication principle fails to solve it? And why does MP fail(?) in the first place?

Three officers—a president, a treasurer, and a secretary— are to be chosen from among four people: Ann, Bob, Cyd, and Dan. Suppose that Bob is not qualified to be treasurer and Cyd’s other commitments make it impossible for her to be secretary. How many ways can the officers be chosen?

There are $14$ ways the officers can be chosen.

No matter how we try, MP can't be applied to this problem? Why?

The objects to be chosen are not interchangeable ("Bob" can't go in the "treasurer" slot while the others can, and "Cyd" can't go in the "secretary" slot while the others can).

If all objects were equally able to go in any slot, you'd have $4\times 3\times 2=24$ ways of choosing, which I assume is the approach you refer to with the phrase "multiplication principle".

• But if we restate the problem as " Three officers—a president, a treasurer, and a secretary—are to be chosen from among four people: Ann, Bob, Cyd, and Dan. Suppose that, for various reasons, Ann cannot be president and either Cyd or Dan must be secretary. How many ways can the officers be chosen? ", we can use multiplication. Can you, please, elaborate on that.
– keys
Commented Apr 25, 2015 at 16:12
• The manner in which you phrase the non-interchangeability (e.g., "alien overlords have decreed that Bob cannot be a secretary") is irrelevant to the mathematics. If not all objects have the same options available to them, you then have to keep track of which objects goes in what slot. Commented Apr 25, 2015 at 16:19
• So, how do we solve the 2nd problem by multiplication? Would "There are three choices for president (all except Ann), three choices for treasurer (all except the one chosen as president), and two choices for secretary (Cyd or Dan). Therefore, by the multiplication rule, there are 3·3·2 = 18 choices in all." be correct?
– keys
Commented Apr 25, 2015 at 16:30
• The point of my answer is that you can't "solve by multiplication". Here's a simpler case: I have two boxes labeled 1 and 2. I have three letters, A, B, and C, and I want to put one letter in each box. If I make no restrictions on what can go where, then my options are (A, B, or C for the first box) $\times$ (whichever two are left for the second box) = $3\times 2$ = $6$. However, if I declare that C cannot go in box 2, then it is incorrect to say "whichever two are left" because sometimes there aren't two left that can go in box two. Commented Apr 25, 2015 at 16:39
• In fact, the possibilities are 1[A] 2[B] (only one option remains if I put A in 1) or 1[B] 2[A] (only one option remains if I put B in 1) or 1[C] 2[A] or 1[C] 2[B] (two options remain if I put C in 1). Commented Apr 25, 2015 at 16:39