Question

How many numbers greater than a million can be formed using the digits 4, 6, 0, 6, 8, 4, 6?

Answer 1

Hint: First, we should know that a million means 7 digits are required. Then we will first find all the possible numbers using all the 7 digits given which will be equal to $\dfrac{7!}{2!3!}$ because we have to divide repeated numbers also. Same way, we have to find numbers starting from 0. After getting this, we will subtract it from total numbers we got earlier to get the final answer.

Complete step-by-step solution -
Here, we know that a million means 1,000,000 i.e. total there are 7 digits required. In the question we are given digits like 4, 6, 0, 6, 8, 4, 6. Out of these, digit 4 is repeated twice and digit 6 is repeated thrice.
Now, we have to find out how numbers can be formed using the digits. So, we will first make 7 blanks i.e. $\_,\_,\_,\_,\_,\_,\_$ . So, in first place we have 7 choices, in second blank we have 6 choices and so on. Thus, we can write it as $7\times 6\times 5\times 4\times 3\times 2\times 1=7!$ . But we know that digit 4 is repeated twice and digit 6 is repeated thrice. So, on dividing this we will get
$=\dfrac{7!}{2!\cdot 3!}=\dfrac{7\times 6\times 5\times 4\times 3!}{2!\cdot 3!}$
On further solving, we will get
$=7\times 6\times 5\times 2=420$
Thus, we can make a total 420 numbers using the digits. But here 0 is there in digit. So, we need to subtract the numbers starting from 0. For this, we will fix 0 in the first blank i.e. $\underline{0},\_,\_,\_,\_,\_,\_$ so now we have 6 blanks left and six digits with us. So, we can write it as $\dfrac{6!}{2!3!}$ ,digit 4 is repeated twice and digit 6 is repeated thrice so dividing this.
On simplification, we get
$\dfrac{6!}{2!3!}=\dfrac{6\times 5\times 4\times 3!}{2!3!}$
$=6\times 5\times 2=60$
So, the number starting from 0 is 60.
We will now subtract this from total 420 numbers and we will get $420-60=360$ numbers.
Thus, we can make a total 360 numbers greater than a million.

Note: Sometimes, students do not divide the repeated digits and lead to false answers. Here, in the given digits, digits 4 and 6 are repeated so, we must divide that number of times in order to get the correct answer. If we simply take $7!=5040$ and then for finding numbers starting from 0, we get $6!=720$ . On subtracting we get $5040-720=4320$ numbers we can make which is wrong because in this many numbers will be repeated. So, we must divide $\dfrac{7!}{2!3!}$ every time we find numbers. So, do not make this mistake.