How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?
Answer
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Hint: In this problem, first we have to find the total possible number with the given digits. Since 2 and 4 are reaped multiple times. So we have to divide the total possible number by the factorial of repeated numbers. Since 0 cannot be in the first place, so we need to calculate all the possible numbers considering 0 at the first place. Finally, we need to subtract that number from the total possible numbers. This will give the required possible number that can formed by the given digit which are greater than 1000000.
Complete step-by-step answer:
Step I
The given digits are 1, 2, 0, 2, 4, 2, 4.
There are a total of 7 digits. Thus 7 digits can be arranged in $7!$ ways.
Here 2 is repeated three times and 4 repeated two times.
Thus 2 can arrange $3!$ ways.
Similarly, 4 can arrange $2!$ ways.
Complete step-by-step answer:
Step I
The given digits are 1, 2, 0, 2, 4, 2, 4.
There are a total of 7 digits. Thus 7 digits can be arranged in $7!$ ways.
Here 2 is repeated three times and 4 repeated two times.
Thus 2 can arrange $3!$ ways.
Similarly, 4 can arrange $2!$ ways.
Step II
Total possible numbers with the given digits
$\begin{array}{l} = \dfrac{{7!}}{{3!2!}}\\ = \dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{3!2!}}\\ = \dfrac{{7 \times 6 \times 5 \times 4}}{2}\\ = 420\end{array}$
Step III
Since 0 cannot be in the first place.
Total number possible when 0 is in the first position
$\begin{array}{l} = \dfrac{{6!}}{{3!2!}}\\ = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!2!}}\\ = \dfrac{{6 \times 5 \times 4}}{2}\\ = 60\end{array}$
Total possible numbers with the given digits
$\begin{array}{l} = \dfrac{{7!}}{{3!2!}}\\ = \dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{3!2!}}\\ = \dfrac{{7 \times 6 \times 5 \times 4}}{2}\\ = 420\end{array}$
Step III
Since 0 cannot be in the first place.
Total number possible when 0 is in the first position
$\begin{array}{l} = \dfrac{{6!}}{{3!2!}}\\ = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!2!}}\\ = \dfrac{{6 \times 5 \times 4}}{2}\\ = 60\end{array}$
Step IV
Subtracting total possible numbers with 0 in the first place from total possible numbers with the given digits, we get the total possible number we can form. Possible number can form
$\begin{array}{l} = 420 - 60\\ = 360\end{array}$
Subtracting total possible numbers with 0 in the first place from total possible numbers with the given digits, we get the total possible number we can form. Possible number can form
$\begin{array}{l} = 420 - 60\\ = 360\end{array}$
Hence, there are total 360 number can be formed by the given digits which are greater than
1000000.
Note: In this question, repetition of numbers is allowed.If the repetition of numbers is not allowed , then the total numbers which can be formed will also be different.So, take care to see in the question if the repetition is allowed or not
1000000.
Note: In this question, repetition of numbers is allowed.If the repetition of numbers is not allowed , then the total numbers which can be formed will also be different.So, take care to see in the question if the repetition is allowed or not
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