Question

Among 10 persons A, B, C are to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is
A. $\dfrac{{10!}}{{24}}$
B. $\dfrac{{9!}}{6}$
C. $\dfrac{{10!}}{6}$
D. None of these

Answer 1

Hint: In the above question, first of all we will select the three positions for A, B and C to speak among 10 persons and then we will arrange the 7 different people to 7 different positions. Also, we will use the concept that selection of r things from n different things is equal to ${}^n{C_r}$.

Complete step-by-step solution -
We have been given that among 10 persons A, B, C are to speak at a function such that A speaks before B and B speaks before C.
We have 10 positions to speak at a function.
So, we have to select the 3 positions for A, B and C according to the given condition.
Selection of a position among 10 positions $= {}^{10}{C_3} \times 1$
Since, the order of A, B and C to speak is fixed. So, its arrangement is equal to 1.
After selecting the 3 positions, we still have 7 positions left for 7 people. So, the arrangement of 7 people to 7 positions is equal to $7!$.
So, total number of ways A, B and C can speak $= {}^{10}{C_3} \times 7!$
$\begin{array}{l} = \dfrac{{10!}}{{7!\,\, \times 3!}} \times 7!\\\ = \dfrac{{10!}}{{1 \times 2 \times 3}}\\\ = \dfrac{{10!}}{6}\end{array}$
Therefore, the correct option is C.

Note: Sometime we just forget that the arrangement of A, B and C are fixed and multiply in arrangement to ${}^{10}{C_3}$ and we get the selection and arrangement of position for A, B and C equal to $\left( {{}^{10}{C_3} \times 3!} \right)$ which is wrong because the order of A, B and C is fixed so its arrangement is equal to 1 only.