permutation basic concept confusion

Question

I have searched on many sites on internet but no one answered my question.. although question is not so different but as i believe in learning concept rather than memorizing it.

I want to know why we multiply in case of permutation.

Like if i want to arrange word "HOME" in different ways my answer will be 4!. My question is why should i multiply 4x3x2x1 to get answer. how multiplication brings me answer. what is the basic concept behind it.

My another question is i tossed two dice and the sum on both faces should be 4 then pair will be (1,3) (3,1) (2,2)

why i cant take (2,2) two times here as it is possible to get this pair two time while tossing a dice in real. Thanks

Answer 1

If you want to make a word of 4 letters using the letters from $\left\{ \mbox{E},\mbox{H},\mbox{M},\mbox{O} \right\}$ (I used a set to emphasize that the order is of no matter at this point), then you have to fill in the four blanks: $$\_ \; \_ \; \_ \; \_$$ For these 'words', the order is of course important. For the first letter, you can still choose any of the four letters. I'll pick $\mbox{O}$, but you could pick any so you have 4 possibilities. At that moment you have: $$\mbox{O} \; \_ \; \_ \; \_$$ and you have to fill the remaining three blanks with the letters from $\left\{ \mbox{E},\mbox{H},\mbox{M} \right\}$. For the second letter, you have 3 possibilities: $$\mbox{O} \; \mbox{H} \; \_ \; \_ \quad , \quad \mbox{O} \; \mbox{E} \; \_ \; \_ \quad \mbox{and} \quad \mbox{O} \; \mbox{M} \; \_ \; \_ $$ But notice that you would also have three (other!) possibilities if we had not started with $\mbox{O}$, but with any of the other three letters. This makes 12 possibilities so far: $4$ for the first letter mulitplied by $3$ for the second letter.

This goes on and can be summarized as: you have

• $4$ possibilities for the first letter, and
• $3$ possibilities for the second letter, and
• $2$ possibilities for the third letter, and
• (only) $1$ possibility (left) for the last letter;

giving a total number of $4 \times 3 \times 2 \times 1 = 4! = 24$ possibilities.

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Another way of visualizing this to understand that you need to multiply, would be drawing a 'tree'. You start from a root and work your way up or down. For the first letter, you already have 4 branches for the 4 possible first letters. Every one of those branches with a fixed first letter, will have 3 branches on the next level with the three possibilities for the second letter; all those 12 branches (because $4 \times 3 = 12$) then split into two branches on the next level for the 2 possibilities to place one of the two remaining letters on the third spot. These 24 possibilities no longer split up, because the last letter will necessarily fill the fourth spot. A lot of text for a simple sketch: draw it!

Answer 2

We multiply because of a property of multiplication; namely that it is distributive over addition. This means that if we multiply two lists or sets $X$ and $Y$, then for every element of $X$, we pair it up with every element of $Y$. And it so turns out that this is exactly the same thing we want to do when we calculate the number of combinations of two events.

So if one event can turn out 2 ways A and B and a 2nd event can turn out 3 ways: 1, 2, and 3, then the number of combinations is 6: A1, B1, A2, B2, A3, B3. We have distributed the elements of one across the elements of the other. And this is the same operation we do (although without realising it most of the time), when we multiply.