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EleoNora [17]
3 years ago
6

Ten friends want to play a game. They must be divided into three teams with three people in each team and one field judge. In ho

w many ways can they do it?
Mathematics
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

<h2> 16,800 number of ways</h2>

Step-by-step explanation:

Let the ten friends represents the 10 letters ABCDEFGHIJ. If they must be divided into three teams with three people in each team and one field judge, the arrangement will become (ABC)(DEF)(GHI)J

This shows that ABC, DEF and GHI are the three teams and J is the chief judge. Since each groups are now a team, we can represent everyone in each teams with the same letter except the judge as shown;

(AAA)(BBB)(CCC)J where J is the judge

Since there are 10 friends in all and there are A, B and C are repeated three times, the arrangement can be done in the following way as shown;

\frac{10!}{3!3!3!1!}

= \frac{10*9*8*7*6*5*4*3!}{3!*6*6*1}\\ = \frac{10*8*7*6*5}{1} \\= 16,800ways

This shows that they can do it 16,800 number of ways

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Step-by-step explanation:

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\left(x^2\sqrt{10}-x\cdot x\sqrt{5}\right)\left(2\cdot x^2\sqrt{15}+x\sqrt{3x}\right)

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10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}

Hence final answer is 10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}


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