Calculate the amount of words that can be made out of ENAHANDA
How many words can be formed by rearranging the letters in ENAHANDA if one requires that the letter D does not have an A directly to the right of it?
I don't understand why the answer is not "The number of all combinations" - (5 over 2) * (6!/2!2!). After all, the rest of the letters could be arranged in 6! ways if we consider that A has to be to the right of D, and since we'd have 2 As and 2 Ns, 6! should be divided by 2!2!. But in the correct solution they seem to skip over the 2!2! part even though we have two double letters.
I don't understand why the answer is not "The number of all combinations" - (5 over 2) * (6!/2!2!). After all, the rest of the letters could be arranged in 6! ways if we consider that A has to be to the right of D, and since we'd have 2 As and 2 Ns, 6! should be divided by 2!2!. But in the correct solution they seem to skip over the 2!2! part even though we have two double letters.
Original post by Nothinghere21
How many words can be formed by rearranging the letters in ENAHANDA if one requires that the letter D does not have an A directly to the right of it?
I don't understand why the answer is not "The number of all combinations" - (5 over 2) * (6!/2!2!). After all, the rest of the letters could be arranged in 6! ways if we consider that A has to be to the right of D, and since we'd have 2 As and 2 Ns, 6! should be divided by 2!2!. But in the correct solution they seem to skip over the 2!2! part even though we have two double letters.
I don't understand why the answer is not "The number of all combinations" - (5 over 2) * (6!/2!2!). After all, the rest of the letters could be arranged in 6! ways if we consider that A has to be to the right of D, and since we'd have 2 As and 2 Ns, 6! should be divided by 2!2!. But in the correct solution they seem to skip over the 2!2! part even though we have two double letters.
Looks like you posted the question here
https://www.reddit.com/r/HomeworkHelp/comments/17e75w9/university_math_how_many_words_can_you_create/
and it was solved?
Original post by mqb2766
Looks like you posted the question here
https://www.reddit.com/r/HomeworkHelp/comments/17e75w9/university_math_how_many_words_can_you_create/
and it was solved?
https://www.reddit.com/r/HomeworkHelp/comments/17e75w9/university_math_how_many_words_can_you_create/
and it was solved?
I don't understand any of the solutions so am looking for alternatives
(edited 6 months ago)
Original post by Nothinghere21
I don't understand any of the solutions so am looking for alternatives
What dont you understand about
1) First, take all the ones that have D at the far right. 7!/2!3!
2) Now there are 7*7!/3!2! left to consider. For everywhere else, before you divide by 2!3!, there are 3 As out of 7 that you don't want to the right of D. So multiply by 4/7. 4*7!/2!3!
3) Add together: 5*7!/2!3!
Original post by Nothinghere21
I don't understand any of the solutions so am looking for alternatives
Just to give you another take on this, you could start with the number of unrestricted words and then subtract the number of words containing the compound letter "DA". This will give you the number of words that do not contain "DA", as required.
8!/(2!3!) - 7!/(2!2!)
Original post by mqb2766
What dont you understand about
1) First, take all the ones that have D at the far right. 7!/2!3!
2) Now there are 7*7!/3!2! left to consider. For everywhere else, before you divide by 2!3!, there are 3 As out of 7 that you don't want to the right of D. So multiply by 4/7. 4*7!/2!3!
3) Add together: 5*7!/2!3!
1) First, take all the ones that have D at the far right. 7!/2!3!
2) Now there are 7*7!/3!2! left to consider. For everywhere else, before you divide by 2!3!, there are 3 As out of 7 that you don't want to the right of D. So multiply by 4/7. 4*7!/2!3!
3) Add together: 5*7!/2!3!
Why do you multiply with 4 in step 2 and 5 in step 3?
Original post by Nothinghere21
Why do you multiply with 4 in step 2 and 5 in step 3?
Ignoring the constraint, there are 7 letters that could go to the right of D. The constraint means 3 aren't allowed (the 3 letter 'A's), so only 4 letters give valid arrangements. So 4/7 of the arrangements ignoring the constraint are going to be valid.
Original post by Nothinghere21
Why do you multiply with 4 in step 2 and 5 in step 3?
Also, in your OP, Im not sure where the 5/2 or 5C2 comes from but you have
3*7*6! = 3*7!
ways of having one of the As after a D, so 6! arrangments of the other lettters, 7 places for a DA and 3 possible As. Then divide by 3!2! to get rid of the repeats. Then
(8! - 3*7!)/2!3!
obviously gives the same result.
(edited 6 months ago)
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