2 of 5 chosen so 1st gets treat, 2nd gets root beer....

Hi everyone,
I still have another question that I can't figure it out. it's about:

2- Two dogs out of 5 are chosen so that the first gets a treat and the second gets a root beer.
a)-How many outcomes are possible?

Thank you so much for your help

Laurie

Re: statistic question #2

You have two "slots" to fill. There are five dogs. So, I guess, any one of the five could be selected to go into the first slot. After that slot is filled, how many choices are left to fill the second slot?
Suppose there were only three dogs.
3 possibilities for the first slot. Then only two possibilities for the second slot. 3*2=6 different ways to fill the two slots. If the dogs are A, B and C, the possibilities are
AB, AC, BA, BC, CA,CB.
Now, can you apply this idea to five dogs?

Re: statistic question #2

Dear Loren,
I did what you explained me, and I found 20 outcomes for my final answer.
thank you so much, Your help has been so helpfull for me

thank you so much

Laurie

Re: statistic question #2

OR:
10 possibilities (treat-beer): 1-2,1-3,1-4,1-5,2-3,2-4,2-5,3-4,3-5,4-5
10 possibilities (treat-beer): 2-1,3-1,4-1,5-1,3-2,4-2,5-2,4-3,5-3,5-4

Re: statistic question #2

lolohelp said:

Hi everyone,
I still have another question that I can't figure it out. it's about:

2- Two dogs out of 5 are chosen so that the first gets a treat and the second gets a root beer.
a)-How many outcomes are possible?

Thank you so much for your help

Laurie

Click to expand...

How many ways can you choose 2 dogs out of 5?

\(\displaystyle {}_{5}C_{2}\, = \, \frac{5!}{(5-2)!\cdot 2!} \, = \, 10\)

How many ways can you give 2 dogs - 2 treats = \(\displaystyle 2\)

How many ways can you choose 2 dogs out of 5 dogs, and , give them two treats = \(\displaystyle 10\cdot 2 \, = \, 20\)