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Hint: To solve the question, we have to apply the divisibility rule of 5 and the formula for selecting the numbers from 2, 3, 5, 6, 7 and 9 to from a third-digit number.

Complete step-by-step answer:

Let the three-digit number be xyz.

The divisibility rule of 5 is the number should have 0 or 5 at the unit’s place.

The value of z can be either 0 or 5.

The given numbers are 2, 3, 5, 6, 7 and 9. Since there is no 0 in the list of given numbers, the value of z is 5.

The values of x, y can be from the set of 2, 3, 6, 7 and 9. Since the numbers are not repeated.

If the value of x is 2 then the value of y can be either of the numbers 3, 6, 7, 9. The case is the same if the value of x is the numbers 3, 6, 7, 9.

Thus, the number of possible ways of selecting x, y values \[={}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}\]

Where \[{}^{5}{{C}_{1}}\] represent the way of selecting one number from the given 5 numbers 2, 3, 6, 7 and 9. \[{}^{4}{{C}_{1}}\] represent the way of selecting one number from the numbers left out after the value of x is chosen.

We know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and \[n!=n\left( n-1 \right)\left( n-2 \right).....2\times 1\]

Thus, the values of \[{}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}\times \dfrac{4!}{1!\left( 4-1 \right)!}\]

\[=\dfrac{5\times 4\times 3\times 2\times 1}{1\left( 4 \right)!}\times \dfrac{4\times 3\times 2\times 1}{1\left( 3 \right)!}\]

\[=\dfrac{5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}\times \dfrac{4\times 3\times 2\times 1}{3\times 2\times 1}\]

\[=5\times 4=20\]

Thus, the number of ways of forming three-digit numbers = 20(way of selecting number 5 for the value of z)

\[=20\times 1=20\]

\[\therefore \]The number of ways three-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 while being no number be repeated = 20

Note: The possibility of mistake can be the calculation since the procedure of solving requires combinations and factorial calculations. The alternative quick can be applying the direct formula of selecting numbers for the three-digit number while keeping the divisibility rule of 5 and condition of numbers being not repeated. This method will lead to the value of answer =\[{}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{1}{{C}_{1}}\]

Complete step-by-step answer:

Let the three-digit number be xyz.

The divisibility rule of 5 is the number should have 0 or 5 at the unit’s place.

The value of z can be either 0 or 5.

The given numbers are 2, 3, 5, 6, 7 and 9. Since there is no 0 in the list of given numbers, the value of z is 5.

The values of x, y can be from the set of 2, 3, 6, 7 and 9. Since the numbers are not repeated.

If the value of x is 2 then the value of y can be either of the numbers 3, 6, 7, 9. The case is the same if the value of x is the numbers 3, 6, 7, 9.

Thus, the number of possible ways of selecting x, y values \[={}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}\]

Where \[{}^{5}{{C}_{1}}\] represent the way of selecting one number from the given 5 numbers 2, 3, 6, 7 and 9. \[{}^{4}{{C}_{1}}\] represent the way of selecting one number from the numbers left out after the value of x is chosen.

We know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and \[n!=n\left( n-1 \right)\left( n-2 \right).....2\times 1\]

Thus, the values of \[{}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}\times \dfrac{4!}{1!\left( 4-1 \right)!}\]

\[=\dfrac{5\times 4\times 3\times 2\times 1}{1\left( 4 \right)!}\times \dfrac{4\times 3\times 2\times 1}{1\left( 3 \right)!}\]

\[=\dfrac{5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}\times \dfrac{4\times 3\times 2\times 1}{3\times 2\times 1}\]

\[=5\times 4=20\]

Thus, the number of ways of forming three-digit numbers = 20(way of selecting number 5 for the value of z)

\[=20\times 1=20\]

\[\therefore \]The number of ways three-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 while being no number be repeated = 20

Note: The possibility of mistake can be the calculation since the procedure of solving requires combinations and factorial calculations. The alternative quick can be applying the direct formula of selecting numbers for the three-digit number while keeping the divisibility rule of 5 and condition of numbers being not repeated. This method will lead to the value of answer =\[{}^{5}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{1}{{C}_{1}}\]

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