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