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If using permutations 6*5*4 would give 120 ways that that dogs could occupy the first, second and third place. Is that correct?

Che_Federer
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  • That's correct. – Théophile May 08 '15 at 20:37
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    I don't understand. Are we not interested in the rest of the runners? For example: Are the following two final arrangements the same $D_1D_2D_3D_4D_5D_6C_1C_2C_3C_4$ and $D_1D_2D_3C_1C_2C_3C_4D_4D_5D_6$? – zoli May 08 '15 at 21:12
  • We are interested in all the runners, so would the number of ways be 144 (120 + 24) if we count the cats as well? – Che_Federer May 08 '15 at 21:31
  • there is no competition. with cats, dogs will always finish first. – abel May 08 '15 at 22:30

1 Answers1

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You're correct that there are $120$ ways dogs can occupy the first three positions if we ignore the rest of the ranking, but as Zoli points out, it's not the full story.

Since we care about all runners, we can first count the number of ways dogs can occupy the first three spots. Like you pointed out, there are $6 \cdot 5 \cdot 4 = 120$ ways this can happen. But we still need to choose an ordering for the remaining $(6 - 3) + 4 = 7$ animals, and you have no restrictions on this ordering.

This is the multiplication principle in action: For a given (valid) ordering of the top $3$, any ordering of the last $10 - 3 = 7$ racers gives us a distinct outcome in which dogs occupy the top $3$ spots. Thus,

$$\text{# of valid outcomes} = 120 \cdot (\text{# of ways to order remaining $7$ animals}).$$

pjs36
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