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17 June, 10:58

In a doctor's waiting room, there are 14 seats in a row. Eight people are waiting to be seated. Three of the 8 are in one family, and want to sit together. How many ways can this happen?

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  1. 17 June, 11:25
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    There are 14 chairs and 8 people to be seated. But among the 8. three will be seated together:

    So 5 people and (3) could be considered as 6 entities:

    Since the order matters, we have to use permutation:

    ¹⁴P₆ = (14!) / (14-6) ! = 2,162,160, But the family composed of 3 people can permute among them in 3! ways or 6 ways. So the total number of permutation will be ¹⁴P₆ x 3!

    2,162,160 x 6 = 12,972,960 ways.

    Another way to solve this problem is as follow:

    5 + (3) people are considered (for the time being) as 6 entities:

    The 1st has a choice among 14 ways

    The 2nd has a choice among 13 ways

    The 3rd has a choice among 12 ways

    The 4th has a choice among 11 ways

    The 5th has a choice among 10 ways

    The 6th has a choice among 9ways

    So far there are 14x13x12x11x10x9 = 2,162,160 ways

    But the 3 (that formed one group) could seat among themselves in 3!

    or 6 ways:

    Total number of permutation = 2,162,160 x 6 = 12,972,960
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