Here Order is important

ie. ** (**1,2,3 **) (**4,5,6**) !=** **(**4,5,6** ) ( **1,2,3**)**

option B doesn't consider this order that's why B is eliminated.

7 votes

An FM radio channel has a repository of $10$ songs. Each day, the channel plays $3$ distinct songs that are chosen randomly from the repository.

Mary decides to tune in to the radio channel on the weekend after her exams. What is the probability that no song gets repeated during these $2$ days?

- $\begin{pmatrix} 10\\ 3 \end{pmatrix}^{2}*\begin{pmatrix} 10\\ 6 \end{pmatrix}^{-1}$
- $\begin{pmatrix} 10\\ 6 \end{pmatrix}*\begin{pmatrix} 10\\ 3 \end{pmatrix}^{-2}$
- $\begin{pmatrix} 10\\ 3 \end{pmatrix}*\begin{pmatrix} 7\\ 3 \end{pmatrix}*\begin{pmatrix} 10\\ 3 \end{pmatrix}^{-2}$
- $\begin{pmatrix} 10\\ 3 \end{pmatrix}*\begin{pmatrix} 7\\ 3 \end{pmatrix}*\begin{pmatrix} 10\\ 6 \end{pmatrix}^{-1}$

1

Here Order is important

ie. ** (**1,2,3 **) (**4,5,6**) !=** **(**4,5,6** ) ( **1,2,3**)**

option B doesn't consider this order that's why B is eliminated.

0

$^{10}C_3$ is the number of ways of choosing 3 songs on day 1. $^7C_3$ is the number of ways of choosing 3 different songs on day 2, so $^{10}C_3$.$^7C_3$ is the number of combinations that meet Mary’s requirement. $^{10}C_3$ $^{10}C_3$ is the total number of ways of choosing 3 songs on each of the two days, without any constraints.

$P=\dfrac{^{10}C_3}{^{10}C_3}\times\dfrac{^7C_3}{^{10}C_3}$

$P=\dfrac{^{10}C_3}{^{10}C_3}\times\dfrac{^7C_3}{^{10}C_3}$

7 votes

**Answer is (C)**

Mary listen to song on weekend $->$ $2$ days

Selecting $3$ songs from $10$ can be done by $10C3$ ways

Now we select $3$ songs from remaining $7$ song in $7C3$

If we select this way we are sure that Mary will listen new song only

Probability can be given by =$\frac{^{10}C_{3}*^{7}C_{3}}{^{10}C_{3}*^{10}C_{3}}$