# Twelve persons are to be arranged around two round tables such that one table can accommodate seven persons and another five persons only. Answer the following questions.

Number of ways in which these 12 persons can be arranged is

(a) $C_{5}^{12}6!4!$

(b) $6!4!$

(c) $C_{5}^{12}6!4!$

(d) None of these

Last updated date: 26th Mar 2023

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There are two round tables such that one can accommodate seven persons and the other table will accommodate five persons.

Now we will find the number of ways five persons can be accommodated in second table out of twelve persons.

As any person can be selected out of \[12\] persons, so number of ways selecting \[5\]persons out of \[12\] persons is,

$C_{5}^{12}$

Now these \[5\] persons can accommodate any seat on the table with \[5\] seats, so number of ways \[5\] person can sit on \[5\] seats of second table is,

$\left( 5-1 \right)!=4!$

So the number of ways \[5\] persons can be accommodated in second table out of twelve persons is,

$C_{5}^{12}4!\ldots \ldots .\left( i \right)$

Once \[5\]persons sit on the second table, there are \[7\] persons left out of \[12\] persons.

So, the number of ways \[7\]persons can seat on first table with \[7\]seats is,

$\left( 7-1 \right)!=6!$

So, the number of ways in which the \[12\] persons can be arranged in two tables is,

$C_{5}^{12}4!\times 6!$

$\Rightarrow C_{5}^{12}6!4!$

Hence, the correct option for the given question is option (a) and (d).

Now we will find the number of ways five persons can be accommodated in second table out of twelve persons.

As any person can be selected out of \[12\] persons, so number of ways selecting \[5\]persons out of \[12\] persons is,

$C_{5}^{12}$

Now these \[5\] persons can accommodate any seat on the table with \[5\] seats, so number of ways \[5\] person can sit on \[5\] seats of second table is,

$\left( 5-1 \right)!=4!$

So the number of ways \[5\] persons can be accommodated in second table out of twelve persons is,

$C_{5}^{12}4!\ldots \ldots .\left( i \right)$

Once \[5\]persons sit on the second table, there are \[7\] persons left out of \[12\] persons.

So, the number of ways \[7\]persons can seat on first table with \[7\]seats is,

$\left( 7-1 \right)!=6!$

So, the number of ways in which the \[12\] persons can be arranged in two tables is,

$C_{5}^{12}4!\times 6!$

$\Rightarrow C_{5}^{12}6!4!$

Hence, the correct option for the given question is option (a) and (d).

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