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!
–
Eli Ross
Staff
·
12 months ago

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@Eli Ross
–
Thank you it is very helpful but i have one more doubt i calculated and answer is coming 3584 is it correct
–
Ishan Dixit
·
12 months ago

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@Ishan Dixit
–
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.
–
Eli Ross
Staff
·
12 months ago

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
Staff
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12 months ago

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@Eli Ross
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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
·
12 months ago

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@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.
–
Eli Ross
Staff
·
12 months ago

## 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! – Eli Ross Staff · 12 months ago

Log in to reply

– Ishan Dixit · 12 months ago

Thank you it is very helpful but i have one more doubt i calculated and answer is coming 3584 is it correctLog in to reply

– Eli Ross Staff · 12 months ago

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.Log in to reply

– Ishan Dixit · 12 months ago

Can i post an image of solution in seperate postLog in to reply

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.)

– Eli Ross Staff · 12 months agoLog in to reply

Log in to reply

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 Staff · 12 months agoLog in to reply

– Ishan Dixit · 12 months ago

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 elseLog in to reply

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. – Eli Ross Staff · 12 months ago

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