- #1

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a) all three digits are different

b) the third digit differs from the first two digits

For (a) I went [tex]1/_9C_3 = 1/84[/tex]. I think it's wrong but I didn't know what else to try.

And for (b) I don't even know where to start.

- Thread starter ms. confused
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- #1

- 91

- 0

a) all three digits are different

b) the third digit differs from the first two digits

For (a) I went [tex]1/_9C_3 = 1/84[/tex]. I think it's wrong but I didn't know what else to try.

And for (b) I don't even know where to start.

- #2

lurflurf

Homework Helper

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In both of these there are 10^3=1000 possibilities soms. confused said:

a) all three digits are different

b) the third digit differs from the first two digits

For (a) I went [tex]1/_9C_3 = 1/84[/tex]. I think it's wrong but I didn't know what else to try.

And for (b) I don't even know where to start.

probability=(# of possibilities that meat criteria)/1000

A) try nPr since order matters.

2) given the two numbers they are written AAB figure how many pairs of numbers are possible.

Last edited:

- #3

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Total possible numbers are 10 and not 9 , count them down if you have any confusion.

Total number of favourable cases are [itex]10 P_3[/itex] and total possible cases are [itex]10^3[/itex].Divide them to get the answer.

For the second one:

Hint: You can fill first two places in anyway you want , that is 10^2 ways , and then the last one in 9 or 8 ways .Add both cases to get the required probability and divide them by / 10^3.

BJ