Answer
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Hint – In this question we have to place some “+” signs and some “-“signs into a circle, so this problem resembles somewhat of a round table arrangement. If we have to place n entities into a round table then there can be $(n - 1)!$ ways to do it so. Use this concept to get the answer.
Complete step by step answer:
As we know that four people cannot sit between three people.
So we have to first arrange five negative (-) signs on a round table.
As we know if there are x candidates arrange on a round table the number of ways is $\left( {x - 1} \right)!$
Therefore the number of ways to arrange five negative (-) signs on a round table is $\left( {5 - 1} \right)!$ = (4!).
Now we have to arrange four positive (+) signs between them so that no two (+) signs are together.
So as we know between five negative (-) signs there are five vacant spaces therefore the number of ways to arrange four (+) signs between five (-) signs so that no two (+) signs are together is (5!).
Therefore the total number of ways to arrange five (-) and four (+) sign so that no two (+) sign are together is
$ \Rightarrow \left( {4!} \right)\left( {5!} \right)$
So, this is the required answer.
Hence option (A) is correct.
Note – Whenever we face such types of problems the key concept that needs to come up into mind is that of the possible arrangements possible into a round table arrangements. Don’t forget to rearrange these signs amongst themselves using the concept of permutation, this will help you get on the right track to get the answer.
Complete step by step answer:
As we know that four people cannot sit between three people.
So we have to first arrange five negative (-) signs on a round table.
As we know if there are x candidates arrange on a round table the number of ways is $\left( {x - 1} \right)!$
Therefore the number of ways to arrange five negative (-) signs on a round table is $\left( {5 - 1} \right)!$ = (4!).
Now we have to arrange four positive (+) signs between them so that no two (+) signs are together.
So as we know between five negative (-) signs there are five vacant spaces therefore the number of ways to arrange four (+) signs between five (-) signs so that no two (+) signs are together is (5!).
Therefore the total number of ways to arrange five (-) and four (+) sign so that no two (+) sign are together is
$ \Rightarrow \left( {4!} \right)\left( {5!} \right)$
So, this is the required answer.
Hence option (A) is correct.
Note – Whenever we face such types of problems the key concept that needs to come up into mind is that of the possible arrangements possible into a round table arrangements. Don’t forget to rearrange these signs amongst themselves using the concept of permutation, this will help you get on the right track to get the answer.
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