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deff fn [24]
3 years ago
9

Of the 10 kids in a chess club, 5 are left-handed and 5 are right-handed. The club holds a round-robin tournament in which every

player plays against every other player exactly once. Of all the matches, how many of them have a left-handed player competing against a right-handed player?
Mathematics
1 answer:
zloy xaker [14]3 years ago
7 0
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The number of ways is equal to 12!8!20!

Step-by-step explanation:

The multiplication principle states that If a first experiment can happen in n1 ways, then a second experiment can happen in n2 ways ... and finally a i-experiment can happen in ni ways therefore the total ways in which the whole experiment can occur are

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Also, given n-elements in which we want to put them in a row, the total ways to do this are n! that is n-factorial.

For example : We want to put 4 different objects in a row.

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Using the multiplication principle and the n-factorial number :

The number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left,all 8 juniors in the middle, and all 20 seniors on the right are : The total ways to put all 12 sophomores in a row multiply by the ways to put the 8 juniors in a row and finally multiply by the total ways to put all 20 senior in a row ⇒ 12!8!20!

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Answer:

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Step-by-step explanation:

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