# There are 9999 tickets bearing numbers 0001, 0002, ……, 9999. If one ticket is selected from these tickets at random, the probability that the numbers on the ticket will consists of all different digits is

(a) $\dfrac{5040}{9999}$

(b) $\dfrac{5000}{9999}$

(c) $\dfrac{5030}{9999}$

(d) None of these

Last updated date: 17th Mar 2023

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Answer

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Hint: To find the probability, we need to find the number of favorable outcomes i.e. the number of ways in which we can draw a ticket having number with different digits on it and also, we need to find the total number of possible cases i.e. the number of ways in which we can draw a ticket from a from 9999 tickets. Both of these can be found by using the concept of permutation and combination.

Before proceeding with the question, we must know all the formulas that will be required to solve this question.

From the number of permutation and combination, the number of ways in which we can select ‘r’ things from total of ‘n’ things is given by,

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.................\left( 1 \right)$

Also, by the concept of principle of counting, if we are given n numbers from which, we have to find the number of 1, 2, 3, 4 - digit numbers having different digits, then it is found by the formula,

$n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right).................\left( 2 \right)$

In this question, we have to select a ticket from 9999 tickets bearing the numbers 0001, 0002, ……, 9999.

Using the formula $\left( 1 \right)$, the number of ways in which we can select 1 ticket from 9999 tickets can be found by substituting n=9999 and r=1 is equal to,

$\begin{align}

& {}^{9999}{{C}_{1}}=\dfrac{9999!}{1!\left( 9999-1 \right)!} \\

& \Rightarrow {}^{9999}{{C}_{1}}=\dfrac{9999\times \left( 9998 \right)!}{\left( 9998 \right)!} \\

& \Rightarrow {}^{9999}{{C}_{1}}=9999.............\left( 3 \right) \\

\end{align}$

Since the tickets are bearing the numbers 0000 to 9999, the number of digits possible in these numbers are equal to 10. Substituting n=10 in formula $\left( 2 \right)$, the number of ways in which a ticket bearing a number with different digits can be drawn from these 9999 tickets is equal to,

\[\begin{align}

& 10\times \left( 10-1 \right)\times \left( 10-2 \right)\times \left( 10-3 \right) \\

& \Rightarrow 10\times 9\times 8\times 7 \\

& \Rightarrow 5040..........\left( 4 \right) \\

\end{align}\]

We know that probability is the ratio of the number of favorable outcomes and the number of outcomes possible. So, using the obtained numbers $\left( 1 \right)$ and $\left( 2 \right)$, we get,

Probability = $\dfrac{5040}{9999}$

Hence, the answer is option (a).

Note: The number of favorable outcomes i.e. the number of ways in which we can select a ticket having a number on it with all different digits can be also found by the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Here, n is the number of possible digits in the number and r is the number of digits in a number. For this question, n=10 and r=4.

Before proceeding with the question, we must know all the formulas that will be required to solve this question.

From the number of permutation and combination, the number of ways in which we can select ‘r’ things from total of ‘n’ things is given by,

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.................\left( 1 \right)$

Also, by the concept of principle of counting, if we are given n numbers from which, we have to find the number of 1, 2, 3, 4 - digit numbers having different digits, then it is found by the formula,

$n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right).................\left( 2 \right)$

In this question, we have to select a ticket from 9999 tickets bearing the numbers 0001, 0002, ……, 9999.

Using the formula $\left( 1 \right)$, the number of ways in which we can select 1 ticket from 9999 tickets can be found by substituting n=9999 and r=1 is equal to,

$\begin{align}

& {}^{9999}{{C}_{1}}=\dfrac{9999!}{1!\left( 9999-1 \right)!} \\

& \Rightarrow {}^{9999}{{C}_{1}}=\dfrac{9999\times \left( 9998 \right)!}{\left( 9998 \right)!} \\

& \Rightarrow {}^{9999}{{C}_{1}}=9999.............\left( 3 \right) \\

\end{align}$

Since the tickets are bearing the numbers 0000 to 9999, the number of digits possible in these numbers are equal to 10. Substituting n=10 in formula $\left( 2 \right)$, the number of ways in which a ticket bearing a number with different digits can be drawn from these 9999 tickets is equal to,

\[\begin{align}

& 10\times \left( 10-1 \right)\times \left( 10-2 \right)\times \left( 10-3 \right) \\

& \Rightarrow 10\times 9\times 8\times 7 \\

& \Rightarrow 5040..........\left( 4 \right) \\

\end{align}\]

We know that probability is the ratio of the number of favorable outcomes and the number of outcomes possible. So, using the obtained numbers $\left( 1 \right)$ and $\left( 2 \right)$, we get,

Probability = $\dfrac{5040}{9999}$

Hence, the answer is option (a).

Note: The number of favorable outcomes i.e. the number of ways in which we can select a ticket having a number on it with all different digits can be also found by the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Here, n is the number of possible digits in the number and r is the number of digits in a number. For this question, n=10 and r=4.

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