Probibility with some identical items

Mathimatician

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A litter of 6 puppies, 3 male and 3 female, is being sold to friendly homes. What is the probability that the first and last puppies sold will be male?

This is my solution:

(4!/3!)/(6!/(3!3!)) = 1/5



the (4!/3!) is the number of ways that you can order the middle 4 puppies assuming that you have a male at front and back (4 puppies to order but 3 are identical).

then the (6!/(3!3!)) is the total number of ways the puppies can be sold.

is this correct?
 
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A litter of 6 puppies, 3 male and 3 female, is being sold to friendly homes. What is the probability that the first and last puppies sold will be male?
This is my solution: 4/(6!/(3!3!)) = 1/5
Is this correct?
\(\displaystyle \dfrac{4}{\dfrac{6!}{(3!)^2}}\) is correct.
 
Hello, Mathimatician!

A litter of 6 puppies, 3 male and 3 female, is being sold to friendly homes.
What is the probability that the first and last puppies sold will be male?

This is my solution:

\(\displaystyle \dfrac{\frac{4!}{3!}}{\frac{6!}{3!3!}} \:=\: \frac{1}{5}\)

The \(\displaystyle \frac{4!}{3!}\) is the number of ways that you can order the middle 4 puppies,
assuming that you have a male at front and back (4 puppies to order but 3 are identical).

Then the \(\displaystyle \frac{6!}{3!3!}\) is the total number of ways the puppies can be sold.

Is this correct? . Yes!

And I like your reasoning, too!

Thank you for showing your work and your thoughts.
 
Of the 20 possibilities, only 4 "make it!":
m m f f f m
m f m f f m
m f f m f m
m f f f m m ...... females have a tendency to get in the way:rolleyes:

I see One male running around three females.

Some will say "__males" are only incomplete females.

Some say Y chromosomes are actually X chromosomes with left-leg broken-off.:eek::eek::eek:
 
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