I think this is a permutations problem. I'm not sure how to go about it.
Problem:
In how many different ways could you select three different days of the week, such that at least one of them begins with the letter "T"? Assume that the order is important.
Is it something like: (7)(6)(5) = 210? Is that the answer? 210 ways?
Thank you for un-confusing me!
Sorry. I misread the OP. I missed the words 'in order'.
There are ways to select any three days of the of the week in order.
There are five days that do not begin with a T.
Therefore, we can select in order which do not begin with a T.
So how many ways can we select three in order so at least one begins with a T?
Well, I can deduce 5 ways, but it's through writing it out, not a formula...
Sunday, Monday, Tuesday ==> 1
Monday, Tuesday, Wednesday ==> 2
Tuesday, Wednesday, Thursday ==> 3
Wednesday, Thursday, Friday ==> 4
Thursday, Friday, Saturday ==> 5
So... i guess it would be (2 [over] 3) = 5?
Hello, Tall Jessica!
In how many different ways could you select three different days of the week,
such that at least one of them begins with the letter "T"?
Assume that the order is important.
How many of these do not begin with a "T"?
For the first day, there are 5 choices: {Sun, Mon, Wed, Fri, Sat}.
For the second day, there are 4 choices.
For the third day, there are 3 choices.
. . Hence, there are: . ways
. . to choose three days that do not begin with "T".
Therefore, there are: . ways to choose three days
. . so that at least one of them does begin with "T".