Natural numbers

Determine the number of all natural numbers greater than 200 in which the digits 1, 2, 4, 6, and 8 occur at most once each.

Correct answer:

n =  288

Step-by-step explanation:

a=4⋅ 4⋅ 3=48 b=5⋅ 4⋅ 3⋅ 2=120 c=5⋅ 4⋅ 3⋅ 2⋅ 1=120  n=a+b+c=48+120+120=288

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Showing 3 comments:
Student
In this question can I get some more explanation pls.

2 years ago  2 Likes
Math student
This is wrong (or the question is written poorly). There are infinite such numbers:

12468
124680
1246800
etc...

2 years ago  2 Likes
Math student
The description is incorrect. The solution above corresponds to the same problem in which there's a limit on the digits in the natural numbers abiding to the given conditions.





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