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The same is true of math problems.
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Four-color lights have six points, so there are either three-color lights or two-color lights.

When there are three lights in one color, first there are four choices for the lights in this color, and then all six lights are arranged in a 6A6 arrangement. Because three of them are of the same color, there is 6A3 arrangement in this three-lamp area, and the number of species in all the six lamps corresponding to seedless arrangement is the same, that is, the above 6A6 arrangement is repeated, and the different arrangement should be 6A6/6A3.

If there are two lights with two colors, the above process is the same. There is 4C2 selection method for lamp selection. Similar to the above arrangement, it can be known that the number of different arrangements of the corresponding lamp selection method is 6A6/(6A2*4A2).

To sum up, the species number is 6A6/6A3*4+6A6/(6A2*4A2)*4C2=36.