How many 3 digit numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6?

Hint: We know that the total number of digits are from 0 – 9 and then we will only take the digits that are not in 0, 2, 3, 4, 5 and 6. And then using the digits we will find all the possible 3 digit numbers by using the method of permutation.
Complete step-by-step answer:
Let’s start our solution,
Permutation means choosing the required number of terms and then arranging them.
Now if we have n different objects and from them we need to pick r objects and then we have to arrange in all possible cases, then the formula is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Now in this case we have to find n.
n will be the number of digits that are not in 0, 2, 3, 4, 5 and 6.
Hence 1, 7, 8, 9.
Therefore we get n = 4
The value of r will be 3, as we need a form 3 digit number only.
Hence, using the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ we get,
$\begin{align}
  & =\dfrac{4!}{\left( 4-3 \right)!} \\
 & =4! \\
 & =24 \\
\end{align}$
Hence, 24 3-digits numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6.

Note:The formula that we have used is very helpful, it helps us to do both choosing and arranging the objects at the same time. One can also solve this question by finding all the possible 3 digits numbers with the given condition and then add them to get the final answer as 24.