How many 4-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5 if the first digit must not be 0 and repetition of digits is not allowed? a.300 b.320 c.360 d.280

Question: How many 4-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5 if the first digit must not be 0 and repetition of digits is not allowed? 

I.

In place of the first digit in the number, we can place 5 digits from the list, since the first digit cannot be 0. We use the placement rule without repetitions.

$A_5^1=n!/(n-m)!=5!/(5-1)!=5$

For the next ones, we do the same, only we take into account the fact that one digit has gone into the first cell, and there are 5 available. We also use the formula placement 3 by 5.

$А_5^3=5!/(5-3)!=3*4*5=60$

Multiply two counted numbers and get the number of possible numbers with a given set of digits.

$A=5*60=300$

Answer: a. 300

II. Another method.

Let's consider each digit of a 4-digit number separately:

 $a_1\not=[0]$ and can take the value $[1,2,3,4,5] \implies a_1=5$

$a_2\not=a_1$ and can take the value $[0,1,2,3,4,5]$ , but $a_1$ is equal to one digit, and under our condition the numbers should not be repeated. $\implies a_2=5$

$a_3\not=a_1\not= a_2$ the same as with $a_2$ , only one less than $a_3=4$

$a_4\not=a_1\not=a_2\not= a_3$ in the same way as with the rest $\implies a_4=3$

Now we multiply all the values of ai.

$a_1*a_2*a_3*a_4=5*5*4*3=300$

Answer: a. 300