document.write( "Question 974075: In how many ways can a family consist of 3 children have different birthdays in a leap year?
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Algebra.Com's Answer #596097 by Edwin McCravy(20064) $\"\"$ $\"About$

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In how many ways can a family consist of 3 children have different birthdays in
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document.write( "There are 366 dates (including Feb.29) the oldest child can have for his/her birthday\r\n" );
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document.write( "For every one of those 366 dates the oldest child can have as his/her birthday,\r\n" );
document.write( "there remain 365 ways the middle child can have a different birthday.\r\n" );
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document.write( "That's 366*365=133590 ways the oldest and middle children can have different\r\n" );
document.write( "birthdays.\r\n" );
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document.write( "For every one of those 366*365=133590 ways the oldest and middle child can have\r\n" );
document.write( "as their birthdays, there remain 364 ways the youngest child can have a\r\n" );
document.write( "different birthday.\r\n" );
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document.write( "That's 366*365*264=133590*364 = 48626760 ways all three children can have\r\n" );
document.write( "different birthdays.\r\n" );
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document.write( "Answer: 48626760.  That's also known as 366P3, \"366 Position 3\", or the\r\n" );
document.write( "number of permutations of 366 things taken 3 at a time.  Order matters\r\n" );
document.write( "because of their birth order, oldest, middle and youngest.\r\n" );
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document.write( "Edwin

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