Question

Find $n\left( S \right)$ for each of the following random experiments.
a) From an urn containing $5$ gold and $3$ silver coins, $3$ coins are drawn at random.
b) $5$ letters are to be placed into $5$ envelopes such that no envelope is empty.
c) $6$ books of different subjects arranged on a shelf.
d) $3$ tickets are drawn from a box containing $20$ lottery tickets.

Answer 1

First, we learn the meaning of the terms permutation and combination which are important topics in probability.
In Probability, the permutation is the process of arranging the outcomes in order. Here, the order must be followed to arrange the items.
In Probability, the term combination refers to the process of selecting the outcomes in which the order does not matter. Here, the order to arrange the items is not followed.
Formula to be used:
The formula to find the permutation is as follows.
${}_n{P_r} = n(n - 1)(n - 2).......(n - r + 1)$
$= \dfrac{{n!}}{{(n - r)!}}$ (! Is a mathematical symbol called the factorial)
Where $n$denotes the number of objects from which the permutation is formed and $r$denotes the number of objects used to form the permutation.
Now, the formula to calculate the combination is as follows.
${}_n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where $n$denotes the number of objects from which the combination is formed and $r$denotes the number of objects used to form the combination.

Complete step by step answer:
Here, we are asked to calculate $n\left( S \right)$ for each of the given random experiments. $n\left( S \right)$ is the required total number of resulting outcomes in the given sample space.
(a) It is given that there are $5$ gold and $3$ silver coins.
So, there are $8$ coins and we need to draw $3$ coins from them.
That is${}_8{C_3}$ ways.
Hence, $n\left( S \right)$ $= {}_8{C_3}$
Using the formula${}_n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we have,
$n\left( S \right) = \dfrac{{8!}}{{\left( {3!} \right)\left( {8 - 3} \right)!}}$
$= \dfrac{{8 \times 7 \times 6 \times 5!}}{{3!5!}}$
$= \dfrac{{8 \times 7 \times 6}}{{3 \times 2}}$
$= 8 \times 7$
$= 56$
Therefore, $3$coins can be drawn in $56$ ways.
(b) Here, we need to use the formula of permutation to calculate $n\left( S \right)$ .
Also,
$n = 5 \\\ r = 5 \\\$
That is ${}_5{P_5}$ ways.
Applying the formula, we have
${}_n{P_r} = n(n - 1)(n - 2).......(n - r + 1)$
${}_5{P_5} = 5(5 - 1)(5 - 2)\left( {5 - 3} \right)\left( {5 - 4} \right)(5 - 5 + 1)$
${}_5{P_5} = 5 \times 4 \times 3 \times 2 \times 1$
$= 120$
Hence, $5$letters are to be placed in $120$ ways.
c)Here, we need to use the formula of permutation to calculate $n\left( S \right)$.
That is ${}_6{P_6}$ ways.
Applying the formula, we have
${}_n{P_r} = n(n - 1)(n - 2).......(n - r + 1)$
${}_6{P_6} = 6(6 - 1)(6 - 2)\left( {6 - 3} \right)\left( {6 - 4} \right)\left( {6 - 5} \right)(6 - 6 + 1)$
${}_6{P_6}_5 = 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$= 720$
Hence, $6$ books on different subjects can be arranged in $720$ ways.
d) So, there are $20$ lottery tickets and we need to draw $3$ tickets from them.
That is ${}_{20}{C_3}$ ways.
Hence, $n\left( S \right)$ $= {}_{20}{C_3}$
Using the formula${}_n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we have,
$n\left( S \right) = \dfrac{{20!}}{{\left( {3!} \right)\left( {20 - 3} \right)!}}$
$= \dfrac{{20 \times 19 \times 18 \times 17!}}{{3!17!}}$
$= \dfrac{{20 \times 19 \times 18}}{{3 \times 2}}$
$= 20 \times 19 \times 3$
$= 1140$
Therefore, $3$ tickets can be drawn in $1140$ ways.

Note: In Probability, the permutation is the process of arranging the outcomes in order. Here, the order must be followed to arrange the items. And the term combination refers to the process of selecting the outcomes in which the order does not matter. Here, the order to arrange the items is not followed.