Question

How many numbers greater than 10 lacs be formed from 2, 3, 0, 3, 4, 2, 3?
(a) 420
(b) 360
(c) 400
(d) 300

Answer 1

Hint:
Here, we will use the fact that all the numbers greater than 10 lacs will not have the digit 0 at the first place. We will use three cases, fixing the three distinct digits other than 0 one by one at the first place, and find the number of ways to arrange the remaining numbers. Then, we will add the results of the three cases to get the required number of ways.
Formula Used: The number of permutations to arrange \[n\] objects is given by \[\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}\], where an object appears \[{r_1}\] times, another object repeats \[{r_2}\], and so on.

Complete step by step solution:
The given digits are 2, 3, 0, 3, 4, 2, 3. Here, 2 is repeated twice, and 3 is repeated thrice.
The 7 digit number greater than 10,00,000 cannot have the digit 0 at the first place. Therefore, the first place can have any of the 3 digits 2, 3, 4.
We will use three cases to find the required answer.
Case 1: The number 2 is the first digit.
In the remaining 6 places, 2 will come 1 time, 0 will come 1 time, 3 will come 3 times, and 4 will come 1 time.
The number of permutations to arrange \[n\] objects is given by \[\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}\], where an object appears \[{r_1}\] times, another object repeats \[{r_2}\], and so on.
The number of ways in which the remaining 6 places can be filled with the remaining digits is given by \[\dfrac{{6!}}{{3!}}\], where 3 is repeated three times.
Therefore, the number of ways to form a 7 digit number greater than 10 lacs from the given digits, where the first digit is 2, is given by
\[\dfrac{{6!}}{{3!}} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} = 6 \times 5 \times 4 = 120\]
Case 2: The number 3 is the first digit.
In the remaining 6 places, 2 will come 2 times, 0 will come 1 time, 3 will come 2 times, and 4 will come 1 time.
Thus, the number of ways in which the remaining 6 places can be filled with the remaining digits is given by \[\dfrac{{6!}}{{2!2!}}\], where 2 and 3 are repeated two times.
Therefore, the number of ways to form a 7 digit number greater than 10 lacs from the given digits, where the first digit is 3, is given by
\[\dfrac{{6!}}{{2!2!}} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 6 \times 5 \times 2 \times 3 = 180\]
Case 3: The number 4 is the first digit.
In the remaining 6 places, 2 will come 2 times, 0 will come 1 time, and 3 will come 3 times.
Thus, the number of ways in which the remaining 6 places can be filled with the remaining digits is given by \[\dfrac{{6!}}{{2!3!}}\], where 2 is repeated two times, and 3 is repeated three times.
Therefore, the number of ways to form a 7 digit number greater than 10 lacs from the given digits, where the first digit is 4, is given by
\[\dfrac{{6!}}{{2!3!}} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}} = 6 \times 5 \times 2 = 60\]
Finally, we will calculate the number of ways in which a number greater than 10 lacs can be formed from 2, 3, 0, 3, 4, 2, 3.
The number of ways in which a number greater than 10 lacs is formed is the sum of the number of ways where the number begins with 2, the number of ways where the number begins with 3, and the number of ways where the number begins with 4.
Therefore, we get the number of ways to form a number greater than 10 lacs can be formed from 2, 3, 0, 3, 4, 2, 3, as \[120 + 180 + 60 = 360\] ways.
Thus, there are 360 numbers greater than 10 lacs that can be formed using the digits 2, 3, 0, 3, 4, 2, 3.

Therefore, the correct option is option (b).

Note:
We can use both combinations and permutations to find the required number of ways in one step.
Any number except 0 can come in the first place.
Therefore, the number of ways to fill the first place \[ = {}^6{C_1}\] ways.
The remaining 6 numbers (including 0) can come in the remaining 6 places.
Thus, the number of ways to fill the remaining places \[ = \dfrac{{6!}}{{2!3!}}\] ways.
Therefore, we get the number of ways to form a number greater than 10 lacs can be formed from 2, 3, 0, 3, 4, 2, 3, as \[{}^6{C_1} \times \dfrac{{6!}}{{2!3!}} = 6 \times \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}} = 360\] ways.
Thus, there are 360 numbers greater than 10 lacs that can be formed using the digits 2, 3, 0, 3, 4, 2, 3.