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**Hint:**The number of ways of selecting $r$ items from $n$ different items is determined by the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Use this formula to find out the number of favorable ways and total number of ways of the given event. The probability is the ratio of them i.e. ${\text{Probability}} = \dfrac{{{\text{No}}{\text{. of favorable ways}}}}{{{\text{Total number of ways}}}}$.

**Complete step by step answer:**

According to the question, there are $5$ duplicates and $10$ original items in an automobile shop.

So the total number of items is $5 + 10$ i.e. $15$. And $3$ items are bought at random from them.

We know that the number of ways of selecting $r$ items from $n$ different items is determined by the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Thus the number of ways of selecting $3$ items from total $15$ items is given as:

$

\Rightarrow {\text{Total number of ways}}{ = ^{15}}{C_3} = \dfrac{{15!}}{{3!12!}} = \dfrac{{15 \times 14 \times 13 \times 12!}}{{6 \times 12!}} \\

\Rightarrow {\text{Total number of ways}} = 455{\text{ }}.....{\text{(1)}} \\

$

Again from the question, all the selected items should be original and not duplicate. Hence, the number of ways of selecting $3$ items from $10$ original items is given as:

$

\Rightarrow {\text{ No}}{\text{. of favorable ways}}{ = ^{10}}{C_3} = \dfrac{{10!}}{{3!7!}} = \dfrac{{10 \times 9 \times 8 \times 7!}}{{6 \times 7!}} \\

\Rightarrow {\text{No}}{\text{. of favorable ways}} = 120{\text{ }}.....{\text{(2)}} \\

$

Further, we know that probability of an event is calculated by the formula:

$ \Rightarrow {\text{Probability}} = \dfrac{{{\text{No}}{\text{. of favorable ways}}}}{{{\text{Total number of ways}}}}$

Therefore, applying this formula and putting the values from equation (1) and (2), we’ll get:

$

\Rightarrow {\text{Probability}} = \dfrac{{120}}{{455}} = \dfrac{{24 \times 5}}{{91 \times 5}} \\

\Rightarrow {\text{Probability}} = \dfrac{{24}}{{91}} \\

$

**$\therefore$ The required probability of selecting $3$ original items from the given items is $\dfrac{{24}}{{91}}$. Hence, option (C) is the correct option.**

**Note:**

The formula used for combination i.e. $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ is just for the number of ways of selecting the items and not for arranging them.

The number of ways of selecting and arranging $r$ items from $n$ different items is given by the formula of permutation and this formula is:

${ \Rightarrow ^n}{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$

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