Use the ideas here: complementary probability, which can be applied to counting (as well as probability).

In particular, there are a total of \(\binom{32}{3}\) combinations of 3 objects.

Now, you need to identify the "bad" choices. The tricky thing is making sure your bad choices are disjoint. Can you think of what might work? (Don't read on before you try to think of it yourself!)

Anyway, there are three kinds of disjoint "bad" choices:

1) All three points are next to each other.
2) Two points are adjacent, but the third isn't.
3) None of the points are adjacent, but two are diametrically opposite.

Try to count the number of choices which would fall into each category!

Could you post your calculations and reasoning for each of the three "bad" cases? Given your answer, I am guessing that you slightly over-counted in Case #2.

If you need to upload the image, you can do that in another post and then copy the link to a comment here using the "Insert an image" button. (Or you could of course just post the picture of your solution in this note.)

@Ishan Dixit
–
You claim that the first point leaves 28 choices, since you can't use the diametrically opposed point or either of the two adjacent points. This is a good observation.

However, you then claim that after the second point is chosen, there are 24 choices for the third point. Is this always true? Or is it possible that, in certain cases, some of the points eliminated by the first point are the same as the ones "eliminated" by the second point, thus leaving more than 24 choices for the third point?

(By the way, these issues you're running into with overlapping is why I suggested the approach with complementary counting with disjoint "bad" choices.)

@Eli Ross
–
I am stuck please provide solution I also found another case in which no. Of group are 32 x 28 x 25 /3!.So should i subtract both cases or do something else

@Ishan Dixit
–
I have laid out three disjoint "bad" choices. Can you try to calculate how many there are of each type?

1) All three points are next to each other.
2) Two points are adjacent, but the third isn't.
3) None of the points are adjacent, but two are diametrically opposite.

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## Comments

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TopNewestUse the ideas here: complementary probability, which can be applied to counting (as well as probability).

In particular, there are a total of \(\binom{32}{3}\) combinations of 3 objects.

Now, you need to identify the "bad" choices. The tricky thing is making sure your bad choices are disjoint. Can you think of what might work? (Don't read on before you try to think of it yourself!)

Anyway, there are three kinds of disjoint "bad" choices:

1) All three points are next to each other.

2) Two points are adjacent, but the third isn't.

3) None of the points are adjacent, but two are diametrically opposite.

Try to count the number of choices which would fall into each category!

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Thank you it is very helpful but i have one more doubt i calculated and answer is coming 3584 is it correct

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Could you post your calculations and reasoning for each of the three "bad" cases? Given your answer, I am guessing that you slightly over-counted in Case #2.

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Use the formatting

`![](image link here)`

If you need to upload the image, you can do that in another post and then copy the link to a comment here using the "Insert an image" button. (Or you could of course just post the picture of your solution in this note.)

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However, you then claim that after the second point is chosen, there are 24 choices for the third point. Is this always true? Or is it possible that, in certain cases, some of the points eliminated by the first point are the same as the ones "eliminated" by the second point, thus leaving more than 24 choices for the third point?

(By the way, these issues you're running into with overlapping is why I suggested the approach with complementary counting with

disjoint"bad" choices.)Log in to reply

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1) All three points are next to each other. 2) Two points are adjacent, but the third isn't. 3) None of the points are adjacent, but two are diametrically opposite.

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