Aptitude Discussion

Q. |
$P$ and $Q$ sit in a ring arrangement with 10 persons. What is the probability that $P$ and $Q$ will sit together? |

✔ A. |
2/11 |

✖ B. |
3/11 |

✖ C. |
2/9 |

✖ D. |
4/9 |

**Solution:**

Option(**A**) is correct

$n(S)=$ number of ways of sitting 12 persons at round table:

$=(12−1)!=11!$

Since two persons will be always together, then number of persons:

$=10+1=11$

So, 11 persons will be seated in $(11−1)!=10!$ ways at round table and 2 particular persons will be seated in $2!$ ways.

n(A)= The number of ways in which two persons always sit together $=10!×2$

\(P(A)=\dfrac{n(A)}{n(S) }\)

\(=\dfrac{10!\times2!}{11!}\)

\(=\dfrac{2}{11}\)

**Joostee**

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**Poonam Pipaliya**

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i don't understand this.

How is it $(12-1)!$?

If seats are 12 then total arrangement would be $12!$, isn't it?

When round tables are considered, we take $(n-1)!$ arrangements for $n$ number of people sitting there. Since on round table there is no start or end point like in a row (where arrangements would be $n!$)

To take an simple example, consider only 2 people sitting around round table. There is only ONE $(=(2-1)!=1!)$ way of sitting there and not $2!=2$ ways.

You may do the math for higher number of people if you wish to.

**Rev**

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There is flaw in the question. As in the 1st statement it says $P$ & $Q$ and in the second statement it says $X$ & $Y$'s probability to sit together.

Kindly correct the question.

No compex maths needed here:

wherever the first person sits in the ring, there are only two sits next to him (right and left).

So the solution is always 2/(N-1) where N is the total number of people in the ring.

Question is a bit misleading, as it actually means 'with 10 OTHER people"