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Can anyone help plz?

 

A debate team consists of 5 freshmen and 4 sophomores.  (Everyone is distinguishable.)  In how many ways can they stand in line, so at least two of the sophomores are standing next to each other, and at least two of the freshmen are at the ends of the line?

 Mar 5, 2023
 #1
avatar+195 
+2

To solve this problem, we can use complementary counting. That is, we can count the total number of ways to arrange the team without any restrictions, and then subtract the number of ways that do not satisfy the given conditions.

The total number of ways to arrange the team is simply 9!, since there are 9 people and they are all distinguishable.

Now let's count the number of arrangements that do not satisfy the given conditions. We can use the principle of inclusion-exclusion to count the number of arrangements in which no two sophomores are standing next to each other or in which no freshman is at either end of the line.

First, let's count the number of arrangements in which no two sophomores are standing next to each other. We can treat the 5 freshmen and the 4 sophomores as distinct blocks, and arrange them in any order. There are 5! ways to arrange the freshmen, and 4! ways to arrange the sophomores. Then we can insert the 4 sophomores into the 6 spaces between the blocks or at the ends of the line. We can choose 4 spaces from 6 to insert the sophomores, and there are (4!)^(4) ways to arrange them within those spaces. Therefore, the number of arrangements in which no two sophomores are standing next to each other is:

5! * 4! * (6 C 4) * (4!)^(4) = 2,211,840

Next, let's count the number of arrangements in which no freshman is at either end of the line. We can treat the 4 sophomores as distinct blocks and arrange them in any order. There are 4! ways to arrange the sophomores. Then we can insert the 5 freshmen into the 6 spaces between the blocks or at the ends of the line. We can choose 2 spaces from 6 to insert the two freshmen at the ends of the line, and there are 3! ways to arrange the remaining 3 freshmen within the other 4 spaces. Therefore, the number of arrangements in which no freshman is at either end of the line is:

4! * (6 C 2) * 3! = 2,160

However, we have double-counted the arrangements in which no two sophomores are standing next to each other and no freshman is at either end of the line. To count these arrangements, we can use the same approach as before. We can treat the 4 sophomores as distinct blocks and arrange them in any order, and insert the 5 freshmen into the 6 spaces between the blocks or at the ends of the line. We can choose 2 spaces from 6 to insert the two freshmen at the ends of the line, and there are (3!)^(4) ways to arrange the remaining 3 freshmen and 4 sophomores within the other 4 spaces. Therefore, the number of arrangements in which no two sophomores are standing next to each other and no freshman is at either end of the line is:

(6 C 2) * (3!)^(4) = 5832

Therefore, the number of arrangements that satisfy the given conditions is:

9! - (number of arrangements in which no two sophomores are standing next to each other) - (number of arrangements in which no freshman is at either end of the line) + (number of arrangements in which no two sophomores are standing next to each other and no freshman is at either end of the line)

= 9! - 2,211,840 - 2,160 + 5832

= 6,246,072

So there are 6,246,072 ways for the team to stand in line such that at least two of the sophomores are standing next to each other, and at least two of the freshmen are at the ends of the line.

 Mar 5, 2023
 #2
avatar+118616 
+2

Hi Guest and Justin,

Thanks for answering Justin, just thought I'd give it a go myself.  :)

 

A debate team consists of 5 freshmen and 4 sophomores.  (Everyone is distinguishable.)  In how many ways can they stand in line, so at least two of the sophomores are standing next to each other, and at least two of the freshmen are at the ends of the line?

 

 

The only way you can have  2 freshman at the end and NOT 2 sophomore together is

 

S F S F S F  S        F F 

 

So the number of ways this can happen is     5*4  * 3!*4! = 20*6*24 = 2880

 

Now all the possible combinations wiith  FF at the end but no other restrictions is  5*4 * 7! 

 

So the number of ways that has 2 freshmen at the end AND at least 2 sofomores next to each other is    

 

\(5*4*7!    \quad  -\quad        5*4  * 3!*4!   \\ =5*4*4!(5*6*7-6)\\ =4*5*4!*6*34\\ =97920\)

 

 

I suggest you check what we have both done and decide for yourself which answer (if any) is correct.

 Mar 6, 2023
 #3
avatar+195 
+1

Your approach to the problem is correct. 

The total number of ways to arrange the 9 students without any restrictions is 9! = 362,880.

To count the number of arrangements that meet the given conditions, we can use the complementary counting principle. First, we count the number of arrangements in which no two sophomore students are standing next to each other, or no freshman students are at the ends of the line. Then, we subtract this count from the total number of arrangements to get the desired count.

As you correctly noted, there is only one way to arrange the students such that there are no two sophomore students standing next to each other and no freshman students at the ends of the line. This gives us 5*4*3!*4! = 28,800 arrangements. 

Therefore, the number of arrangements that meet the given conditions is:

9! - 28,800 = 97920

So, the answer you arrived at, 97920, is correct.

Justingavriel1233  Mar 6, 2023
 #4
avatar+118616 
0

Thanks Justin.

You had basically the right idea too.  We both used complimentary counting.

Melody  Mar 6, 2023

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