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Saumyojit

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Three variants of exam paper are to be given to 12 students. In how many ways can the students be placed in 2 rows of 6 each so that there should be no identical variants side by side and that the students sitting one behind the other should have the same variant. Find the number of ways it can be done

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Suppose one row -> __ ___ ___ ____ ___ ___ = first seat has 3 options(VAr 1 ,2 ,3) , second seat has two options (var 1, var 2) assume var 3 has already been placed in first seat , .... 2 options from the second seat to Fifth , last one has one option .
SO expression is (3c1 * 2c1* 2c1 *2c1 * 2c1) = 48

Then 48 * 12 c6 * 6! *6!
WHere am i worng?
 
Saumyojit said:
Three variants of exam paper are to be given to 12 students. In how many ways can the students be placed in 2 rows of 6 each so that there should be no identical variants side by side and that the students sitting one behind the other should have the same variant. Find the number of ways it can be done

Suppose one row -> __ ___ ___ ____ ___ ___ = first seat has 3 options(VAr 1 ,2 ,3) , second seat has two options (var 1, var 2) assume var 3 has already been placed in first seat , .... 2 options from the second seat to Fifth , last one has one option .
SO expression is (3c1 * 2c1* 2c1 *2c1 * 2c1) = 48

Then 48 * 12 c6 * 6! *6!
WHere am i worng?
Click to expand...
Your error is that the number of possibilities for the later seats depends on choices made earlier. If both copies of variant A have been used, then you'll only have one choice for the one after variant B, not two.

Here's how I'd approach it:

First, clearly the two rows are identical, so we can restate the problem as asking for the number of ways to arrange AABBCC with no adjacent letters identical. But the interrelationships, as I said above, make that complicated.

To get a feel for how it will work, I'll just make a list of arrangements starting with AB:

ABACBC (ABAB... can't work)​
ABCABC​
ABCACB​
ABCBAC​
ABCBCA​

Having done this, I see that we can make all other arrangements by permuting ABC, which can be done in 6 ways. For example, permuting to BAC swaps A and B, changing ABCBCA to BACACB, and so on.

So it looks like the answer is 6*5 = 30.

Now, in order to get an answer anything like the options given, I have to suppose that what is being done ("the number of ways this can be done") includes both placing the exams on the desks and placing students at the desks, which is not quite clear in the problem. The latter can be done in 12! ways. So the answer would then be 30*12! . Looking at the options, 12C6*6!*6! = 12!. So (d) is the answer.
 
Yes you have written down the 5 arrangmenets but then i dont understand WHy 6 *5 =30 ?
Dr.Peterson said:
Having done this, I see that we can make all other arrangements by permuting ABC, which can be done in 6 ways. For example, permuting to BAC swaps A and B, changing ABCBCA to BACACB, and so on.

So it looks like the answer is 6*5 = 30.
Click to expand...
 
Saumyojit said:
Yes you have written down the 5 arrangmenets but then i dont understand WHy 6 *5 =30 ?
Click to expand...
I think you mean, why I multiplied by 6, not why 6 times 5 equals 30!

I said, "Having done this, I see that we can make all other arrangements by permuting ABC, which can be done in 6 ways."

What that means is that each of my 5 arrangements produces 5 more when you replace A, B, and C, respectively, with some other order of the letters, such as B, A, C. I gave one example. I could, if you wish, list all 30 possibilities, making 5 rows of 6, but that should not be necessary.
 
Dr.Peterson said:
What that means is that each of my 5 arrangements produces 5 more when you replace A, B, and C, respectively, with some other order of the letters, such as B, A, C.
Click to expand...
Well, i understood that those 5 arrangement each produces another five adding to 30 .
But what I was rather asking is why did you write only 5 arrangement in first place when you could have wirtten more .
Why did you wrote those 5 only ? whats the pattern u are following? You also could have written another 5 only instead of those 5 .

Dr.Peterson said:
you replace A, B, and C,
Click to expand...
ABCBCA this will give another 5 . But what is the pattern of replacing and rearranging.
U gave this ABCBCA to BACACB what i see is that u changed b and a . WHat will the other 4 of this ABCBCA
 
Saumyojit said:
Why did you wrote those 5 only ? whats the pattern u are following?
Click to expand...
As I said, "I'll just make a list of arrangements starting with AB"; there are no others. The list was arranged in terms of the next letter, which could only be A or C, and so on.
Saumyojit said:
ABCBCA this will give another 5 . But what is the pattern of replacing and rearranging.
U gave this ABCBCA to BACACB what i see is that u changed b and a . WHat will the other 4 of this ABCBCA
Click to expand...
As I said, "we can make all other arrangements by permuting ABC, which can be done in 6 ways. For example, permuting to BAC swaps A and B, changing ABCBCA to BACACB, and so on." The permutation BAC means A becomes B, B becomes A, and C is unchanged:

ABC​
BAC​

Other permutations work similarly.

I assume you eventually understood this, but we may as well help others understand.
 
Form a decision tree, if first 4 people are given AB, CA (comma denotes the rows) the next 2 should be: A,B;A,C or C,B; Then all 3 sub decisions can have 3 sub decisions… and 3 sub decisions can have another 3 sub decisions, till 3^4. Depend on the number of the ways of how to hand out to the first 4 students, n*3^4 is number of ways. That’s how I would approach this and it seems that there are no double counting.
 
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Dr.Peterson said:
As I said, "I'll just make a list of arrangements starting with AB"; there are no others. The list was arranged in terms of the next letter, which could only be A or C, and so on.

As I said, "we can make all other arrangements by permuting ABC, which can be done in 6 ways. For example, permuting to BAC swaps A and B, changing ABCBCA to BACACB, and so on." The permutation BAC means A becomes B, B becomes A, and C is unchanged:

ABC​
BAC​

Other permutations work similarly.

I assume you eventually understood this, but we may as well help others understand.
Click to expand...
My understanding is with restriction lifted there are 3^12 ways in handing out exam papers and to reduce it to 30 it’s a bit hard to see. Can you point out the flaws in my approach?
 
sky2rain said:
My understanding is with restriction lifted there are 3^12 ways in handing out exam papers and to reduce it to 30 it’s a bit hard to see. Can you point out the flaws in my approach?
Click to expand...
Which restrictions are you ignoring? What is your final answer?

I think you are ignoring the fact that "each variant is used for four students" (seen in the image of the problem, but not copied in the text of the OP).
 
Dr.Peterson said:
Which restrictions are you ignoring? What is your final answer?

I think you are ignoring the fact that "each variant is used for four students" (seen in the image of the problem, but not copied in the text of the OP).
Click to expand...
Sorry, didn’t see that
 
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