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Sagot :
Answer:
The answer is "840".
Step-by-step explanation:
Following are the number of ways in which selecting a team A by 9 children:
[tex]= ^{9_{C_{3}}\\\\\\[/tex]
[tex]=\frac{9!}{3! \times 6!} \\\\=\frac{9\times 8\times 7\times 6!}{3 \times 2\times 1\times 6!}\\\\=\frac{9\times 8\times 7}{3 \times 2\times 1}\\\\=\frac{3\times 4\times 7}{1}\\\\=\frac{84}{1}\\\\=84[/tex]
Following are the number of ways in which selecting a team B by remaining 6 children:
[tex]= ^{6}_{C_{3}}[/tex]
[tex]= \frac{6!}{(3! \times 3!)}\\\\= \frac{6!}{(3\times 2\times 1 \times 3!)}\\\\= \frac{6\times 5 \times 4 \times 3!}{(3\times 2\times 1 \times 3!)}\\\\= \frac{ 5 \times 4 \times 3!}{3!}\\\\= 5 \times 4 \\\\=20[/tex]
Following are the number of ways in which selecting a team C by remaining 3 children:
[tex]= ^{3}_{C_{3}}\\\\=\frac{3!}{3!}\\\\= 1[/tex]
Following are the number of ways in which making 3 teams by 9 children:
[tex]= \frac{(84 \times 20 \times 1)}{3!}\\\\= \frac{(84 \times 20 )}{6}\\\\= 14 \times 20\\\\= 280\\\\[/tex]
(Note: we've split by 3! Because it also is necessary to implement three teams between themselves)
Now 3 leagues have to be played. One is going to be run by each team.
That is the way it is
Different possible divisions
[tex]= 280 \times 3!\\\\= 280 \times (3 \times 2 \times 1)\\\\= 840[/tex]
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