We want to paint three rooms in a house, each a different color, and we may choose from seven different colors of paint. How many color combinations are possible for the three rooms?

Question

We want to paint three rooms in a house, each a different color, and we may choose from seven different colors of paint. How many color combinations are possible for the three rooms?

Answer 1

Let the 7 colors be red, orange, yellow, green, blue, white, and violet.

The first color can be any one of 7.

The second color can be any one of 6.

The third color can be any one of 5.

Write 7×6×5 equal to 42×5 or 21×10 or 210 possible color permutations for the three rooms. Note that, in a permutation, selections of red-blue-orange, red-orange-blue, blue-red-orange, blue-orange-red, orange-red-blue, & orange-blue-red are all counted as possible outcomes of a selection of 3 out of 7 colors even though all 6 outcomes have the same three colors in different orders.

The number of permutations is also obtained by writing ₇P₃ which translates to 7!/(7−3)! or (7×6×5×4×3×2×1)÷(4×3×2×1) or 210 as above.

In a count of combinations, however, all but one of these 6 outcomes would be discarded so that one would calculate (7×6×5) − (7×5×5) equal to (7×5) times (6 − 5) or 35 in order to find the number of possible combinations of 3 colors chosen (without replacement in the "hat") from 7 colors.

This is also obtained by writing ₇C₃ which translates to 7!/(3!(7-3)!) or (7×6×5×4×3×2×1)÷(3×2×1)÷(4×3×2×1)

which reduces to (7×6×5)÷(3×2×1) or 35 combinations as above.