Question

How many numbers greater than a million can be formed by using the digits 7, 4, 6 and 0 if 4 has to be​

Answer 1

Any number of greater than a million will contain all the seven digits.

Now, we have to arrange these seven digits, out of which 2 occurs twice, 3 occurs twice and the rest are distinct.

The number of such arrangements =

2!×3!

7!

=

2!×3!

7×6×5×4×3!

=7×6×5×2=420

These arrangements also include those numbers which contain 0 at the million’s place.

Keeping 0 fixed at the millionth place, we have 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are distinct.

These arrangements also include those numbers which contain 0 at the million’s place.

Keeping 0 fixed at the millionth place, we have 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are distinct.

These 6 digits can be arranged in

2!×3!

6!

=

2!×3!

6×5×4×3!

=6×5×2=60 ways

Hence, the number of required numbers =420–60=360.

(please mark the answer as a brain list)

Answer 2

Answer:

A total of 360 numbers is greater than a million.

Step-by-step explanation:

Given:

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 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 the total numbers we got earlier to get the final answer.

To find:

Series

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 7, 6, 0, 6, 8, 4, and 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 the first place, we have 7 choices, in the second blank we have 6 choices, and so on. Thus, we can write it as $7*6*5*4*3*2*1=7!$ . But we know that digit 4 is repeated twice and digit 6 is repeated thrice. So, on dividing this we will get

$=7!2!3!\\=7*6*5*4*3!2!*3!$

On further solving, we will get

$=7*6*5*2=420$

Thus, we can make a total of 420 numbers using the digits. But here 0 is there the in the digit. So, we need to subtract the numbers starting from 0. For this, we will fix 0 in the first blank i.e. 0–,_,_,_,_,_,_ so now we have 6 blanks left and six digits with us. So, we can write it as 6!2!3! digit 4 is repeated twice, and digit 6 is repeated thrice so dividing this.

On simplification, we get

$6!2!3!=6*5*4*3!2!3!\\=6*5*2=60$

So, the number starting from 0 is 60.

We will now subtract this from a total of 420 numbers and we will get $420-60=360$ numbers.

Thus, we can make a total of 360 numbers greater than a million.

Note: Sometimes, students do not divide the repeated digits, which leads to false answers. Here, in the given digits, digits 4 and 6 are repeated so, we must divide that number of times 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 $7!2!3!$ every time we find numbers. So, do not make this mistake.