What are the possible arrangements of BENNETT TAN with separated T's and N's?

[elitewarr]

Homework Statement

Find the number of different arrangements of the name "BENNETT TAN" with all the T's separated and all the N's separated.

Homework Equations

The Attempt at a Solution

I have no idea how to start this question.

If anyone is so kind to help me. Thank you. I need some urgent help!

 

[tiny-tim]

hi elitewarr!

i think you'll have to do it by subtraction …

find all the ways without the restriction, then subtract all the ways that have two Ts or Ns together

(i expect it'll be really messy :yuck:)

 

[Robert1986]

Homework Statement

Find the number of different arrangements of the name "BENNETT TAN" with all the T's separated and all the N's separated.

Homework Equations

The Attempt at a Solution

I have no idea how to start this question.

If anyone is so kind to help me. Thank you. I need some urgent help!

A very important problem solving technique is to break the problem into manageable chunks, and then put the pieces together. This is how I will help you solve your problem.

First, here is the way that I interpreted your problem: How many ways can you arrange the letters of BENNETT TAN so that there are no two consecutive T's and no two consecutive N's. So, BENNETT TAN is not a legal arrangement, but BENTNTETAN is legal. Is that correct?

First, I would solve this problem: How man ways can you arrange the letters: BEETT TA so that there are no two consecutive T's. To solve this problem, I use something I call the "gap method". Consider the following diagram:

$ # $ # $ # $ # $

In this diagram, the #'s represent positions of the letters B,E,E, and A and the $'s represent legal positions for the T's. Now, each of the #'s are going to be used, but clearly not all of the $'s are going to be used. Does this make sense so far? If not, think about it some. If you don't understand something, just ask. So, now just look at the #'s and the letters B,E,E,A. There are absolutley no restrictions on how these can be placed on the #'s, right? Well, how many ways can you arrange the letters B,E,E,A? This is the same as the number of ways you can position the letters on the #'s. Now you need to count the number of ways that you can position the T's. There are 5 $'s which means there are 5 legal places to put the three T's. How many ways can you place three T's in 5 positions? Multiply these two numbers together and you have the total number of ways that you can arrange the letters BEETT TA such that there are no consecutive T's. Am I being clear? Don't hesitate to ask questions if you need to do so. Now, let's deal with the N's. So far, we have constructed a string from the letters BEETTTA such that there are no consecutive T's. Now, we want to add in the N's. Note that there is no way that we can add an N to make the T's illegal. Consider this diagram:
$#$#$#$#$#$#$#$
where the #'s are the letters we have already placed and the $'s are legal positions for the T's. Now, how many ways can you place your T's. Multiply this number by the number in the first part and you have your answer. Note that you don't have to do a bunch of messy subtraction, either.

For practice, solve this problem (which is easier than the current one):
How many ways can you arrange the letters in COMBINATORICS so that there are no two consecutive vowels?

 

[elitewarr]

Thanks for the reply guys!

So for me to arrange the 4 letters BEEA, there will be 4! / 2! ways.
After which, since there are 5$, I will be choosing 3 $ out of the 5 to place the T's. So, it will be 5C3.

Now if we put the 3 N's into the newly-formed string, there will be 8C3 ways.

So in total, there are 4!/2! * 5C3 * 8C3

Am I right??

Thanks again :D

 

[Robert1986]

Thanks for the reply guys!

So for me to arrange the 4 letters BEEA, there will be 4! / 2! ways.
After which, since there are 5$, I will be choosing 3 $ out of the 5 to place the T's. So, it will be 5C3.

Now if we put the 3 N's into the newly-formed string, there will be 8C3 ways.

So in total, there are 4!/2! * 5C3 * 8C3

Am I right??

Thanks again :D

Perfect! However, and this is merely a matter of aesthetic preferences on my part, I would write 4!/2! like this:
4C2 * 2C1 * 1C1

Since it reflects the idea that first you find 2 spots for the E's, then one spot for the B and then one spot for the A. But, like I said, that it merely my personal preference, you got the problem correct, and, better yet, your solution fits on one line instead of having to use a lot of messy subtraction.

The technique that we used to solve this can be used to solve just about any problem of the "arrange letters so that such and such happens" variety. Did you try the COMBINATORICS problem I gave you for practice?

 

[elitewarr]

For the problem you gave, I will be inserting the vowels O, I and A into the spaces between other letters. Like, $#$#$#$#$#$#$#$#$, where $ is the possible ways I can slot O,I and A and # are the rest of the letters. So first I arrange the #, which is 8! / 2!. For the vowels, it will be 9C5. Total ways = 8!/2! * 9C5. Am i right??

Thanks a lot robert! :D

 

[Robert1986]

For the problem you gave, I will be inserting the vowels O, I and A into the spaces between other letters. Like, $#$#$#$#$#$#$#$#$, where $ is the possible ways I can slot O,I and A and # are the rest of the letters. So first I arrange the #, which is 8! / 2!. For the vowels, it will be 9C5. Total ways = 8!/2! * 9C5. Am i right??

Thanks a lot robert! :D

You are correct about the consonants, but you have treated the vowels as if they are all the same. So, you have 9 legal spots for the vowels, you need to determine how many ways you can place the O's, then how many ways you can place the I's then how many ways you can place the A.

 

[elitewarr]

Oh... So if I changed my answer to 8!/2! * 9C2 * 7C2 * 5C1, where 9C2 is to select ways to place O's, 7C2 to place I and 5C1 to place A, am I right??

Thanks!