5 BOYS AND 5 GIRLS SIT IN A ROW AT RANDOM. THE PROBABILITY THAT THE BOYS AND GIRLS SIT ALTERNATIVELY
Problem on Classical definition.
Answers
Method - 1:
Given that 5 boys and 5 girls sit in a row at random.
Total number of ways in which 10 persons can be arranged in a row = 10!.
= > 5 boys can be arranged themselves in 5! ways.
1 B 2 B 3 B 4 B 5 B 6
Now,
Then there are 6 positions left(1,2,3,4,5,6).
The 5 girls can be seated in (1,2,3,4,5) (or) (2,3,4,5,6) = > 2 ways.
The girls can be arranged themselves in 5! ways.
Therefore, the number of ways both girls and boys can sit alternatively = 2 * 5! * 5!
Hence,
Required probability = 2 * 5! * 5!/10!
= > 2 * 5 * 4 * 3 * 2 * 1 * 5 * 4 * 3 * 2 * 1/10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= > 2 * 5 * 4 * 3 * 2 * 1/ 1 0 * 9 * 8 * 7 * 6
= > 4/9 * 8 * 7
= > 4/504
= > 1/126.
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Method 2:
= > Total number of arrangement of 5 boys and 5 girls = 10!.
Boys can be seated alternatively in 5! ways.
Girls can be seated alternatively in 5! ways.
This arrangement can be started in 2 ways( A girl can start (or) A boy can start).
= > Required probability = 2 * 5! * 5!/10!
= > 1/126.
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Hope this helps!