Question

Ten guests are to be seated in a row of which three are ladies. The ladies insist on sitting together while two of the gentlemen refuse to take consecutive seats. In how many ways can the guests be seated?

Answer 1

Collect the data which was in the given question. Here, the arrangement is important so we use the permutation concept or we can use the counting concept. Counting technique is the number of ways to choose $k$ objects from a group of $n$ objects. General form of choosing $k$ objects from a group of $n$ objects is $$k!$$$$$$. Permutation is a collection or a combination of objects from a set where the order or the arrangement of the chosen object does matter.

Formula used: General formula for permutation chosen $r$ things from $n$ objects ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$

Complete step-by-step answer:
It is given that the ten guests are seated in a row
Number of guests is $10$ and number of seats is $10$
Out of ten, three are ladies
Ladies is $3$ members
Gents is $7$ members
The ladies insist on sitting together
$3$ ladies of them is sit together
$2$ gents are refused to consecutive seats
To find the number of way guest to seated
Condition \left( 1 \right)$$$$3 ladies of them are sitting together
Let the girls be ${G_1},{G_2},{G_3}$ are sit together
$\left( 2 \right)$, $2$ gents are refused to consecutive seats
Let the gents be ${B_1},{B_2}$ are not sit together
The ten seats are
\underline {} $$$$\underline {} $$$$\underline {} $$$$\underline {} \underline {} $$$$\underline {} \underline {} $$$$\underline {} $$$$\underline {} $$$$\underline {}
We are going find
“The difference of total number of seating $7$ gents, $3$ ladies together and the number of ways of two gents with three girls sit together and six remaining gents together is”
From that, we get answers.
The three seats considered as one sits by condition $\left( 1 \right)$ ${G_1}, {G_2}, {G_3}$ seats
\underline {} $$$$\underline {} $$$$\underline {} $$$$\underline {} \underline {} $$$$\underline {} $\underline {}$ $\underline {{G_1}{G_2}{G_3}}$ considered as one seat
The boys are ${B_1},{B_2},{B_3},{B_4},{B_5},{B_6},{B_7}$
${B_1}{B_2}{B_3}{B_4}{B_5}{B_6}{B_7}$ $\underline {{G_1}{G_2}{G_3}}$
Here the total number of seats is eight seats and one seat has three ladies is
$\Rightarrow$ $8!3!$
The number of ways of two gents seating together is $2!$
The number of ways of two gents with three girls together an five remaining gents is
Seven seats, one seat three ladies sit together, another seat two gents sit together, remaining $5$ gents are sit together is
$\Rightarrow$ $7! \times 3! \times 2!$
Therefore the difference is $8! \times 3! - 7! \times 3! \times 2!$
$\Rightarrow$ $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$
$\Rightarrow$ $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$
$\Rightarrow$ $3! = 3 \times 2 \times 1 = 6$
$\Rightarrow$ $2! = 2$
That is we can write it as,
$\Rightarrow$ $40320 \times 6 - 5040 \times 6 \times 2$
On multiplying the terms,
$\Rightarrow$ $241920 - 60480$
On subtracting the terms,
$\Rightarrow$ $181440$
Hence the number of ways of $10$ guests can sit is $325440$

Note: We solve this sum by counting method, by using permutation the same answer will get. The only way to solve this type sum is to understand the given data is more important.