Permutations and Combinations

[bubbagump]

In how many different ways can a panel of 12 jurors and 2 alternates be chosen from a group of 18 prospective jurors?

Sportywarbz had a similar post but I can't figure it out based on their question...
Can you get me started on the right path?

The first jurror can be chosen in 18 ways

The second jurror can be chosen in 17 ways ? The first & the second jurror can be chosen in (18*17) ways

and continue....

 

[soroban]

Hello, bubbagump!

In how many different ways can a panel of 12 jurors and 2 alternates be chosen from a group of 18 prospective jurors?

\(\displaystyle \text{First, choose 12 jurors from the available 18 people.}\)
. . \(\displaystyle \text{There are: }\:{18\choose12} \:=\:\frac{18!}{12!\,6!} \;=\;18,\!564 \text{ ways.}\)

\(\displaystyle \text{Then choose 2 alternates from the remaining 6 people.}\)
. . \(\displaystyle \text{There are: }\:{6\choose2} \:=\:\frac{6!}{2!\,4!} \:=\:15\text{ ways.}\)

\(\displaystyle \text{Therefore, there are: }\:18,\!564 \times 15 \:=\:278,\!460\text{ ways.}\)


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Another approach:

Partition the 18 people into three groups:
. . 12 who will be jurors,
. . 2 who will be alternates, and
. . 4 who will not serve.

\(\displaystyle \text{There are: }\:{18\choose12,2,4} \:=\: \frac{18!}{12!\,21\,4!} \;=\;278,\!460\text{ ways.}\)