Koty's Dad or Math Guru

[barry_virus]

Extra Credit Problem for a Stats Course I am in:

There are 23 students in a class :

11 boys 12 girls. The teacher assigns seats at random . How many diff boy/girl seating assignments are possible?


A gumball machine had 7 red and 5 yellow gumballs left. You put in the pennies until you empty the machine. How many diff sequences of red and yellow gumballs can come out of the machine?

Mark can remember only the first 3 digits of his friend's phone number. He also knows that the number has 7 digits and that the last digit is not a 0. If he dialed all of the possible no's and if it takes him 24 sec per try, how long would it take him to do every possibility?

:scared

 

[ocelot]

These 2 problems have exactly the same solution technique being really the same kind of problem I believe.

I'm thinking about it.

 

[ocelot]

Sorry, I should have said the 1st two problems are the same kind. The 3rd is a little harder.

 

[ocelot]

The gumball problem is tricky because it phrases the problem as the number of different "sequences" of red and yellow which makes one want to think the answer involves number of permutations rather than number of combinations but since there is no difference between one red ball and another or one yellow ball and another the problem really is one of the number of combinations.

This involves the combinatorial formula: [ N! / (n!) (N-n)! ]

So for problem 2 if looked at from the point of view of red gumballs: [ 12! / ( 7! 5! ) ]

From yellow point of view [ 12! / ( 5! 7! ) ] which of course are the same and equal 792

 

[ocelot]

Use same idea for problem 1.

 

[ocelot]

Last problem IS a permutation problem. It boils down to how many possible 3-digit sequences are there (choosing from 10 possible digits). Once this is known just multiply by 9 since for every possible 3-digit sequence there are 9 possible 4th and final digit.

Once this is found simply multiply by 24 secs per permutation.

Use the 10 choosing 3 permutation formula [ N! / n! ] the number of possible 3-digit sequences = 10! / 3!

Good luck.

 

[barry_virus]

Thank You Ocelot!

Very much apprecaited!!! :mj14:

 

[KotysDad]

Use same idea for problem 1.

Be careful. The two problems arent the same.

After you find the places for the boys or girls, the boys and girls arent identical as in the gumballs.

You have to multply in a factor of 11! and 12! since the boys and girls themselves can be rearranged.

 

[KotysDad]

The last problem is just 10*10*10*9 * 24 sec

 

[ocelot]

Ouch I bow to you KotysDad. I do believe you are correct. I should not have spoken. Very sorry BV!

 

[barry_virus]

:mj14: I bow to both of you!!! :mj14: