How many 3 - digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Question

Prepp Team · Accepted Answer

Finding 3-Digit Numbers Divisible by 5

We need to form 3-digit numbers using the digits 2, 3, 5, 6, 7, and 9. The numbers must meet two conditions:

They must be divisible by 5.
None of the digits can be repeated.

Understanding Divisibility by 5

A number is divisible by 5 if its unit digit (the rightmost digit) is either 0 or 5. In the given set of digits {2, 3, 5, 6, 7, 9}, the only digit that can be the unit digit for a number to be divisible by 5 is 5.

Forming the 3-Digit Number

Let the 3-digit number be represented by three positions: Hundred's place, Ten's place, and Unit's place.

We will fill the positions based on the conditions:

Unit's Place: The digit must be 5 for the number to be divisible by 5.
Number of options for the unit's place = 1 (the digit 5).
Hundred's Place: The digit 5 has been used for the unit's place. We have 6 digits in total {2, 3, 5, 6, 7, 9}. After using 5, the remaining digits are {2, 3, 6, 7, 9}. There are 5 remaining digits.
Number of options for the hundred's place = 5 (any of the digits 2, 3, 6, 7, 9).
Ten's Place: We have used one digit for the unit's place (5) and one digit for the hundred's place from the remaining 5 digits. This means we have used 2 digits in total from the original set of 6 digits. The number of remaining digits is $6 - 2 = 4$.
Number of options for the ten's place = 4 (any of the remaining 4 digits).

Calculating the Total Number of Numbers

To find the total number of 3-digit numbers that satisfy the conditions, we multiply the number of options for each place:

Total numbers = (Options for Hundred's Place) $\times$ (Options for Ten's Place) $\times$ (Options for Unit's Place)

Total numbers = $5 \times 4 \times 1 = 20$

Thus, 20 three-digit numbers can be formed from the digits 2, 3, 5, 6, 7, and 9, which are divisible by 5 and have no repeated digits.

Revision Table: Steps to Form Numbers Divisible by 5 (No Repetition)

Place	Condition/Constraint	Digits Available	Number of Choices
Units	Must be 5 (for divisibility by 5)	{5}	1
Hundreds	Cannot be the digit used in Units place	{2, 3, 6, 7, 9} (Original set excluding 5)	5
Tens	Cannot be the digits used in Units or Hundreds place	Remaining 4 digits (Original set excluding 5 and the digit used in Hundreds)	4
Total Number of Numbers			$1 \times 5 \times 4 = 20$

Additional Information: Permutations and Combinations

This problem involves the concept of permutations because the order of the digits matters (e.g., 235 is different from 325). Specifically, it's a permutation problem with constraints.

Permutation: The arrangement of objects in a specific order. The number of permutations of $n$ distinct objects taken $r$ at a time is denoted by $P(n, r)$ or $_nP_r$, and is calculated as $P(n, r) = \frac{n!}{(n-r)!}$.
Combination: The selection of objects where the order does not matter. The number of combinations of $n$ distinct objects taken $r$ at a time is denoted by $C(n, r)$ or $_nC_r$, and is calculated as $C(n, r) = \frac{n!}{r!(n-r)!}$.

In our problem, we fix one position (units) and then arrange the remaining digits in the other positions. The selection of digits for the hundreds and tens place from the available pool, followed by their arrangement in specific places, falls under the principle of counting arrangements, which is related to permutations.

If there were no repetition and no divisibility constraint, the number of 3-digit numbers from 6 distinct digits would be $P(6, 3) = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{720}{6} = 120$. The constraints significantly reduce the possibilities.