Finite Math Permutation: Logic Flaw: An interesting scenario that we have 4 diff

Question

Finite Math

Permutation: Logic Flaw:

An interesting scenario that we have 4 different spots and we need to fill them with 2 colors while each color occupy 2 spots. And let assume we have 3 colors to choose from. We know the right way to do is C(3, 2) × P(4, 2). But somehow some logic like, OK, let me pick 2 spots, then pick 1 color for those two boxes and then pick another color to fill the rest 2 boxes, the computation for that is C(4, 2) × C(3, 1) × C(2, 1). This is an overcount case. Can you explain where the overcount happens?

Explanation / Answer

Let the three colours be R,B,G.

Let the spots be A,B,C,D

According to the flawed logic ; Consider the case when

I first pick two boxes (A,B) and then I pick the colour R . Rest of the two boxes (C,D) be coloured with G.

This case is again counted when I pick the other two boxes (C,D) and the colour G to colour them and let the rest of the boxes (A,B) be coloured with R.

Thus we have overcounted this case.

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.