`person( 1 )` starts counting at `START`, and `he( 1 )` counts by `cardinal( INCR )`s.

If `START` is the 1st number that `person( 1 )` counts. what is the `NTH`th number that `he( 1 )` counts?

`ANSWER`

What is the first number that `he( 1 )` counts?

`START`

What is the second number that `he( 1 )` counts?

`\begin{align*}&`

`START` + `INCR` \\
&= `START + INCR`\end{align*}

What is the third number that `he( 1 )` counts?

`\begin{align*}&`

`START` + `INCR` + `INCR` \\
&= `START` + (2 \times `INCR`) \\
&= `START + 2 * INCR`\end{align*}

What is the `NTH`th number that `he( 1 )` counts?

`\begin{align*}&`

`START` + (`NTH - 1`\times`INCR`) \\
&= `START` + `( NTH - 1 ) * INCR` \\
&= `ANSWER`\end{align*}

There are `NUM` people in a room.

If everyone shakes everyone else's hand exactly once, how many handshakes occurred?

`ANSWER`

Given `NUM` people, each person shakes the hands of `NUM - 1` other people.

The following is almost the answer.

`NUM` \times `NUM - 1` = `NUM * ( NUM - 1 )`

We have double counted the handshakes though, since `person( 1 )` shaking `person( 2 )`'s hand is the same handshake as `person( 2 )` shaking `person( 1 )`'s hand.

Therefore, the following is the correct answer.

`\dfrac{`

`NUM` \times `NUM - 1`}{2} = `ANSWER`

`person( 1 )` wants to give `his( 1 )` friend a potted plant. At the local florist, the flowers come in `NUM_COLORS` colors, and there are `NUM_POTS` types of flower pots.

If `he( 1 )` can choose any flower and any pot, how many different potted plants can `person( 1 )` buy?

`ANSWER`

If `person( 1 )` decided on a flower color, how many different potted plant combinations are there?

`He( 1 )` can choose one of `NUM_POTS` flower pots, and so there are `NUM_POTS` different potted plants possible (given that `he( 1 )` already chose a flower color).

Since there are `NUM_COLORS` flower colors, there are

possible potted plants.`NUM_COLORS` \times `NUM_POTS` = `ANSWER`