medicoremathstudent
New member
- Joined
- Jun 16, 2021
- Messages
- 7
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"?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 \(\dbinom{6}{3}=20~\&~\dbinom{3}{2}=3\)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:
Please share your work/thoughts about this problem
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.
7200... but I still don't understand how to think of the solution myself. How do you know to multiply (20)(3)(5!)?\((20)(3)(5!)=~?\)
I understand why I would need to do 6C3 x 3C2, but I don't quite grasp why that is multiplied by 5!.7200... but I still don't understand how to think of the solution myself. How do you know to multiply (20)(3)(5!)?
................different arrangements.........
Pay attention to the word arrangements in your OP.
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.