Question

How many numbers greater than a million can be formed with the digits $2,3,0,3,4,2,3$?

Answer 1

Hint: One millions contains seven digits that is, $1$ millions = $10,00,000$ .
Therefore, any number greater than a million has seven digits.
We have to arrange seven digits $2,3,0,3,4,2,3$. The number of ways in which seven digits is arranged is $7!$ and since number $2$ occurs twice and the number $3$ is occurred thrice so we divide $7!$ by $2! \times 3!$
$ \Rightarrow \dfrac{{7!}}{{2! \times 3!}}$
This includes the digit $0$at the million’s place.
Use the factorial:
$n!$ = 1 $\times$ 2 $\times$ 3$\times$ 4 $\times$ ……..$n$
Next, we find the number of ways of arranging the digits where $0$is fixed at the millionth place. Because we have to find numbers greater than a million.
 There are remaining $6$ out of which 2 occurs twice, 3 occurs thrice.
These six digits can be arranged in $\dfrac{{6!}}{{2! \times 3!}}$.
The required number that is greater than a million is $\dfrac{{7!}}{{2! \times 3!}} - \dfrac{{6!}}{{2! \times 3!}}$.

Complete step-by-step answer:
Here, one million = $10,00,000$. One million contains seven digits. The numbers that are greater than $10,00,000$ can be arranged by seven numbers $2,3,0,3,4,2,3$.
The number greater than $10,00,000$ has seven places out of which 2 occurs twice, 3 occurs thrice and the rest are distinct
A total number of ways arranging seven digits =$\dfrac{{7!}}{{2! \times 3!}}$.
Use the factorial: $n! = 1 \times 2 \times 3\times 4 \times ……..n$
Total number of ways arranging seven digits =$\dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{2! \times 3!}}$.
$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{2! \times 3!}} = \dfrac{{7 \times 6 \times 5 \times 4}}{2}$
$ \Rightarrow \dfrac{{7 \times 6 \times 5 \times 4}}{2} = \dfrac{{840}}{2}$
$ \Rightarrow \dfrac{{840}}{2} = 420$
These arrangements also contain the number $0$ at the million’s place, so we have to keep $0$ at the millionth place.
The number starts with $0$ is, not a seven-digit number.
There are remaining $6$ numbers, out of which 2 occurs twice, 3 occur thrice.
These six digits can be arranged in $\dfrac{{6!}}{{2! \times 3!}} = \dfrac{{6 \times 5 \times 4 \times 3!}}{{2! \times 3!}}$
$ \Rightarrow \dfrac{{6 \times 5 \times 4 \times 3!}}{{2! \times 3!}} = \dfrac{{6 \times 5 \times 4}}{2}$
$ \Rightarrow \dfrac{{6 \times 5 \times 4}}{2} = 60$
The number of required numbers is a total number of ways arranging seven digits subtracted by the arrangement of $6$ digit numbers.
The required numbers of arrangement: $420 - 60 = 360$.

Final Answer: The numbers greater than a million can be formed with the digits $2,3,0,3,4,2,3$ are $360$.

Note:
The factorial is the multiplication of consecutive terms from one to . Factorial, gives the number of ways in which objects can be permuted
$n!$ = 1 $\times 2 \times 3\times 4 \times ……..n$
A student may forget to find the numbers of the arrangement of the $6$ digit numbers and that has to be subtracted by the number of ways of arranging seven digits.
The place values:
The place value for ten lakhs or million is \[10,00,000\].
The place value for one lakh is \[1,00,000\].
The place value for 10 thousand is \[10,000\].
The place value for 1 thousand is \[1,000\].
The place value for 1 hundred is \[100\].
The place value for tens is \[10\].
The place value for ones is \[1\].