Combinatorics question

[medicoremathstudent]

3 letters are selected from the word WINTER and 2 letters are selected from the word DAY. The number of different arrangements of the five letters is...

 

[Deleted member 4993]

3 letters are selected from the word WINTER and 2 letters are selected from the word DAY. The number of different arrangements of the five letters is...

How many letters are there in "WINTER"?

How many ways you can select 3 letters from there?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[pka]

3 letters are selected from the word WINTER and 2 letters are selected from the word DAY. The number of different arrangements of the five letters is...

Note that there is no overlap in the two words \(\dbinom{6}{3}=20~\&~\dbinom{3}{2}=3\)
How many sets of five does that make? Each of those can be rearranged in \(5!\) ways.

 

[medicoremathstudent]

How many letters are there in "WINTER"?

How many ways you can select 3 letters from there?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

My bad. I tried 6P3 x 3P2 and got 720 ways. But the answer is 7200 ways.

 

[medicoremathstudent]

Note that there is no overlap in the two words \(\dbinom{6}{3}=20~\&~\dbinom{3}{2}=3\)
How many sets of five does that make? Each of those can be rearranged in \(5!\) ways.

So, the number of different ways I can select 3 letters from"winter" is 20, and there are 3 ways to select 2 letters from "day". 23 letters make 4.6 sets of 5. Do I do 4.6x5! then?

 

[pka]

\((20)(3)(5!)=~?\)

 

[medicoremathstudent]

\((20)(3)(5!)=~?\)

7200... but I still don't understand how to think of the solution myself. How do you know to multiply (20)(3)(5!)?

 

[medicoremathstudent]

7200... but I still don't understand how to think of the solution myself. How do you know to multiply (20)(3)(5!)?

I understand why I would need to do 6C3 x 3C2, but I don't quite grasp why that is multiplied by 5!.

 

[Deleted member 4993]

Pay attention to the word arrangements in your OP.

................different arrangements.........

 

[medicoremathstudent]

Pay attention to the word arrangements in your OP.

I don't understand, why do (20)(3)(5!) instead of taking (20x3)P5?

 

[Cubist]

I don't understand, why do (20)(3)(5!) instead of taking (20x3)P5?

20*3 is the number of ways that:- ( 3 letters can be selected from the word winter AND 2 letters can be selected from the word day )
That gives 60 different sets of 5 letters.

But then you need to think how many ways that the 5 letters in EVERY set could be arranged. IF you had 3 letters total then you could arrange them in the following ways { abc, acb, bac, bca, cab, cba } which is 3! = 6 ways. There are 3 ways you can select the first letter, 2 ways you can select the second letter, and then there's one letter left (3*2*1). I'm not willing to list all the ways that 5 letters can be arranged but hopefully you can see that it would be 5! = 120 ways.

 

[medicoremathstudent]

20*3 is the number of ways that:- ( 3 letters can be selected from the word winter AND 2 letters can be selected from the word day )
That gives 60 different sets of 5 letters.

But then you need to think how many ways that the 5 letters in EVERY set could be arranged. IF you had 3 letters total then you could arrange them in the following ways { abc, acb, bac, bca, cab, cba } which is 3! = 6 ways. There are 3 ways you can select the first letter, 2 ways you can select the second letter, and then there's one letter left (3*2*1). I'm not willing to list all the ways that 5 letters can be arranged but hopefully you can see that it would be 5! = 120 ways.

This really cleared things up, thank you very much!