Question

Eight persons amongst whom A,B and C are to speak at a function. Number of different ways can it done if A wants to speak before B and B wants to speak before C is
(a) $3360$
(b) $13440$
(c) $5040$
(d) $6720$

Answer 1

To solve this question, we will use permutation and combination. Since, eight places are given and three among them will be arranged and their order matters. So, we will use combination formulas for them. Now, there are five places left to arrange but their order does not matter. So, we will use permutation for them and multiply with combinations to get the required answer.

Complete step by step solution:
The given condition is that there are a total of eight members to participate and all of them will participate but the three of them will be arranged in an order.
So, let’s first do the calculation for three among them that have been arranged in an order. Thus, we will us the formula of combination as:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\cdot \left( n-r \right)!}$
Now, we will substitute $8$ for $n$ and $3$ for $r$ as:
$\Rightarrow {}^{8}{{C}_{3}}=\dfrac{8!}{3!\cdot \left( 8-3 \right)!}$
Here, we will solve the bracketed term first and we will expand the numerator also as:
$\Rightarrow {}^{8}{{C}_{3}}=\dfrac{8\times 7\times 6\times 5!}{3!\cdot 5!}$
Now, we will cancel out the equal like terms as:
$\Rightarrow {}^{8}{{C}_{3}}=\dfrac{8\times 7\times 6}{3!}$
Here, we will expand the denominator and will do require calculation as:
$\begin{aligned} & \Rightarrow {}^{8}{{C}_{3}}=\dfrac{8\times 7\times 6}{3\times 2\times 1} \\\ & \Rightarrow {}^{8}{{C}_{3}}=\dfrac{8\times 7\times 6}{6} \\\ \end{aligned}$
Now, after cancelling the equal like term and doing multiplication, we will have:
$\begin{aligned} & \Rightarrow {}^{8}{{C}_{3}}=8\times 7 \\\ & \Rightarrow {}^{8}{{C}_{3}}=56 \\\ \end{aligned}$
Since, we got the arrangement of three persons. Now, we will do the calculation for remaining five persons without any order using permutation as;
$\Rightarrow 5!=5\times 4\times 3\times 2\times 1$
After doing multiplication, we will get:
$\Rightarrow 5!=120$
Now, the required number of ways $={}^{8}{{C}_{3}}\times 5!$
After substituting the corresponding values, we will get:
The required number of ways $=56\times 120$
Now, we will do multiplication as:
The required number of ways $=6720$
Hence, the option D is the correct answer.

Note: Permutation is the way of choosing any objects from all the given objects where the order of chosen objects does not matter but the combination is the way of choosing objects and the order of choosing objects is important.