Question

First, second and third prizes are to be awarded at an engineering fair in which 13
exhibits have been entered. In how many different ways can the prizes be awarded?​

Answer 1

Answer:

13p 3 ways=1716 ways

Step-by-step explanation:

13 fact/3 fact=1716 ways

Answer 2

Answer:

First, second and third prizes can be awarded to 13 exhibits in 1716 different ways.

Step-by-step explanation:

There are total 13 participants or exhibits who will be awarded first, second, and third prizes.

The first prize can distributed to 1 participant out of 13 participants in $\[^{13}{C_1}\]$ different ways.

Now, 12 participants are remaining for being awarded.

So, the second prize can distributed to 1 participant out of 12 participants in $\[^{12}{C_1}\]$ different ways.

Now, 11 participants are remaining for being awarded.

So, the third prize can distributed to 1 participant out of 11 participants in $\[^{11}{C_1}\]$ different ways.

Now, using the fundamental theorem of arithmetic for multiplication,

First, second and third prizes can be distributed to 13 participants in $\[^{13}{C_1}{ \times ^{12}}{C_1}{ \times ^{11}}{C_1}\]$ different ways.

Therefore,

$\[^{13}{C_1}{ \times ^{12}}{C_1}{ \times ^{11}}{C_1}\]=13\times12\times11\\\[^{13}{C_1}{ \times ^{12}}{C_1}{ \times ^{11}}{C_1}\]=1716$

Therefore, first, second and third prizes can be distributed to 13 participants in 1716 different ways.