1. ## Permutations (I think...)?

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!

2. Originally Posted by Tall Jessica
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?
That is the number of ways to select any three days of the week in order.
The number of ways to select three days of the week in order none of which begins with a T is $(5)(4)(3)$
So what is the answer for at least one?

3. Originally Posted by Plato
That is the number of ways to select any three days of the week in order.
The number of ways to select three days of the week in order none of which begins with a T is $(5)(4)(3)$
So what is the answer for at least one?
Hm... I'm not sure why (5)(4)(3) would be three days that don't begin with a "T"... BUT, my guess would be (3)(2)(1)?

If that's the right answer, I'm a little lost! Excuse my lack of mathematical genius )

4. Sorry. I misread the OP. I missed the words 'in order'.
There are $\binom{7}{3}=35$ 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 $\binom{5}{3}=10$ 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?

5. 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?

6. Originally Posted by Tall Jessica
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?
There are $25$ ways to select three days of the week such that at least one begins with a T.

7. 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.

$\text{There are: }\:_7P_3 \:=\:\dfrac{7!}{4!} \:=\:210 \text{ ways to choose }any\text{ three days.}$

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: . $5\cdot4\cdot3 \,=\,60$ ways
. . to choose three days that do not begin with "T".

Therefore, there are: . $210 - 60 \:=\:150$ ways to choose three days
. . so that at least one of them does begin with "T".

8. Oh, wow.

But, HOW do I get to that answer?