Question

The number of different signals that can be sent with three dots and two dashes are
A. 5
B. 6
C. 10
D. 20

Answer 1

If there are n things to be arranged in a row, among which a things are of one kind, b things are of another kind, and c things are of another, then the total number of arrangements is given by $\dfrac{{n!}}{{a!b!c!}}$, by using this formula we will get the required solution.

Complete step by step Answer:

Given that there are three dots and two dashes first.
Therefore, there are 5 elements in total, where 3 of them are of one kind, and two of them are of another kind.
Therefore the number of ways these dots and dashes can be arranged in a line is:
$= \dfrac{{5!}}{{3!2!}}$
On simplification we get,
$= \dfrac{{120}}{{6 \times 2}}$
On multiplying the denominator we get,
$= \dfrac{{120}}{{12}}$
On cancelling the common factors we get,
$= 10$
Therefore, the number of different signals that can be sent with three dots and two dashes is 10.
Hence, option (C) is the correct one.

Note: A permutation is an act of arranging the objects or numbers in order while Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.

The formula for permutations is given by: ${}^n{P_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{\left( {n - r} \right)!}}$
The formula for combinations is given by: ${}^n{C_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Repetitions in the arrangement: Note that if there are n things to be arranged in a row, among which a things are of one kind, b things are of another kind, and c things are of another, then the total number of arrangements is given by $\dfrac{{n!}}{{a!b!c!}}$.