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# There are six balls in a box. Three balls are marked with 1,

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There are six balls in a box. Three balls are marked with 1, [#permalink]

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07 Aug 2003, 01:44
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There are six balls in a box. Three balls are marked with 1, three with 0. Three balls are taken at random without repetitions. In how many combinations will the product of the digits on the balls equal to 0?
Manager
Joined: 24 Jun 2003
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07 Aug 2003, 02:40
The product of three numbers will equal ‘0’ if at least one of the numbers is zero. Another way is to substract a number of combinations when none of the numbers equal zero from a total number of combinations.

A total number of combinations of 3 balls out of 6 is 6C3=20. The number of combinations of all ones is 3C3=1. Thus, the number of combinations of none zero out of three is 20-1=19.

Double check the longer way (at least)

Number of combinations of exactly one zero (two '1'): 3C1*3C2=9
Number of combinations of exactly two zeros (one '1'): 3C2*3C1=9
Number of combinations of exactly tree zeros: 3C3=1

- Stolyar?
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Respect,

KL

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07 Aug 2003, 10:42
Guys,

We are paying so much attention to combinations here. It is great, but I must say that another topic - promultation is a kind of deep forest for me.

If someone has any questions on promultation that he would like to post here, I would be grateful.
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Respect,

KL

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07 Aug 2003, 12:52
Saylor

Are you sure the answer is 19. We didn't have 6 distinct balls.
We two type of three balls. Then the combination must be much
smaller than 19.

I don't have mathametical expression for the problem.
But here is my logical explanation.

Following are the possible combinations:
111
110
100
000

Because it is combinaiton problem the oder is not important.
So, 110 and 101 are one at the same.

SVP
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08 Aug 2003, 01:21
N(product is zero)=N(total)-N(product is 1)

N(total)=6C3=20
N(product is 1)=3C3=1

20-1=19

what is wrong? I use combinations. Where order is not important.
08 Aug 2003, 01:21
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# There are six balls in a box. Three balls are marked with 1,

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