Question

How many numbers greater than $10,00,000$can be formed by using the digits $1$,$2$,$0$,$2$,$4$,$2$,$4$ ?

Answer 1

the given questions revolve around the concepts of permutations and combinations. We have to find the numbers of numerals that can be formed using the digits given to us in the problem itself $1$,$2$,$0$,$2$,$4$,$2$, $4$. There are various ways to do the task like using the concept of counting through multiplication and concept of combinations. Then, we will also have to eliminate the numbers that are equal or less than $10,00,000$ and arrive at our final answer.

Complete step by step answer:
Number of digits given to us $= 7$
The digits given to us are: $1$,$2$,$0$,$2$,$4$,$2$, $4$.
So, the number of times $0$ appears in the digits$= 1$
Number of times $1$ appears in the digits$= 1$
Number of times $2$ appears in the digits$= 3$
Number of times $4$ appears in the digits$= 2$
Now, we have to form a $7$ digit number as the number we make must be greater than $10,00,000$. So, the first digit of the number from left at ten lakhs place value must be greater than or equal to zero as the number formed must be greater than $10,00,000$.
So, the options for a digit at ten lakhs place value are: $1$,$2$,$4$.
Now, if the digit at the tens lakh place is $1$, we are left with digits $2$,$0$,$2$,$4$,$2$, $4$.
Now, we can arrange them in any order as the number formed would be greater than $10,00,000$ in all cases. So, the number of numerals that can be formed in this case are $\dfrac{{6!}}{{2!\, \times \,3!}} = \dfrac{{6 \times 5 \times 4}}{2} = 60$.
Now, if the digit at the tens lakh place is $2$, we are left with digits $1$,$0$,$2$,$4$,$2$, $4$.
Now, we can arrange them in any order as the number formed would be greater than $10,00,000$ in all cases. So, the number of numerals that can be formed in this case are $\dfrac{{6!}}{{2!\, \times \,2!}} = \dfrac{{6 \times 5 \times 4 \times 3}}{2} = 180$.
Now, if the digit at the tens lakh place is $4$, we are left with digits $1$,$0$,$2$,$2$,$2$, $4$.
Now, we can arrange them in any order as the number formed would be greater than $10,00,000$ in all cases. So, the number of numerals that can be formed in this case are $\dfrac{{6!}}{{3!}} = 6 \times 5 \times 4 = 120$.
Hence, the total number of numerals that can be formed with the digits $1$,$2$,$0$,$2$,$4$,$2$, $4$ and are greater than $10,00,000$ are $60 + 180 + 120 = 360$

Note: For solving the given question, we need to understand the basic concepts of permutations and combinations. We must know how to calculate the number of permutations when the entities given are repeating also.