How do I solve this math problem?

Answer 1

There are a total of ? ways of painting the cards.

Each of the four cards can be painted four ways. There's no rules against them all being the same color or all different colors, so the only thing that matters is that there are four cards and four colors. So the first can be any of four colors, and for each of those the second card can be any of four colors, so on and so forth.

4 * 4 * 4 * 4 = 256 total ways of painting the cards.

There are ? ways of painting them using all four colors.

Let's assume that the order we paint them in doesn't matter. The first card can be painted 4 colors. The second card can't be painted an already used color, so there are only 3 options remaining. The next card only has two options, and the last card has 1.

4 * 3 * 2 * 1 = 24 ways of painting the cards using every color.

There are ? ways of painting two cards red, one card black, and one card blue.

We're going to work a little backwards on this one, since the relevant information is which card is given to which color. So two cards need to be red, but crucially the first one can be any card, and the second one only has three remaining options. The black card can be any of the two remaining cards, and the blue card has only one choice left at that point (no matter which card we choose for black).

Again, the equation that we come up with 4 * 3 * 2 * 1, although we arrived at it from a different angle.

Effectively, there will always be 24 ways of creating a specific combination of colors, we've just arbitrarily changed which specific set we're selecting for.

There are ? ways of painting the four cards using three colors.

The first card can be any of the four colors. The second card can be any of three colors, not matching the first. The third can can be any remaining color. Since we have four cards and three colors, the last card must match one of the cards we've already painted, and therefore the last card has 3 choices.

4 * 3* 2 * 3 = 72

There are ? ways of painting them using two colors.

Similar to the last one, the first card can be any of four, but we'll make the second one a different color. The last two cards can each be either of the colors we've already used, so they each have two options.

4 * 3 * 2 * 2 = 48

Answer 2

You can solve mathematics problems, by going to your teacher to show you on how to calculate or discuss with your friends, who know how to calculate mathematics and they pass mathematics.