how many numbers divisible by 5 formed by 0 1 2 3, digits can't repeat.
Answers
Answer:660
Step-by-step explanation:
Possible 5-digit numbers formed which are divisible by 5 will be of 2 types: those ending with 0 and those ending with 5.
In the case of numbers ending in 0, we can choose the other 4 digits in 6P4 ways, using permutations since the order of the choices matters.
6P4 = 6!/(6–4)!
= 6!/2!
= 720/2 = 360 ways.
In the case of numbers ending in 5, we can choose the other 4 digits in 6P4 ways minus 5P3 ways (since 5P3 of these 6P4 choices will start with 0 making them invalid 5-digit numbers). Fixing 5 as the last digit and 0 as the first digit will leave 5P3 choices for the middle 3 digits. Hence the subtraction.
6P4 - 5P3
= 6!/(6–4)! - 5!/(5–3)!
= 6!/2! - 5!/2!
= 360 - 60
= 300 ways.
Therefore, total possible 5-digit numbers divisible by 5 can be formed using the digits 0, 1, 2, 3, 4, 5, 6 without repetition is:
360 + 300
= 660.
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