Question

How many numbers between 2000 and 3000 can be formed from the digits 2, 3, 4, 5, 6 and 7 when repetition of digits is not allowed?

Answer 1

Hint: We will use the formula of permutation which is given by $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ where n are the total number of objects and r is the selected objects. We will use this formula to find the four digits number in which repetition is not allowed. As this formula is applied only for the questions in which there is no repetition.

Complete step-by-step answer:
Now we consider here a 4 digit number as $\underline{{}}\,\underline{{}}\,\underline{{}}\,\underline{{}}$. We have total digits as 6 out of which we have to form the four digit number. The digits are given to us as 2, 3, 4, 5, 6 and 7. As we are here to find the numbers between 2000 and 3000, one thing is clear here that the first digit is going to be 2 only. Therefore, we have first place filled by 1 choice. Thus, we get $\underline{1}\,\underline{{}}\,\underline{{}}\,\underline{{}}$. We are supposed to now find out the rest of places filled up by how many choices. For this we will use the concept of permutation. The formula of permutation is given by the formula $P_{r}^{n}=\dfrac{n!}{\left( n-r \right)!}$ where n is the total number of objects and r is the selected objects. As we are left with 3, 4, 5, 6 and 7 therefore, we have n = 5 and r = 3. This results into
$\begin{align}
  & P_{3}^{5}=\dfrac{5!}{\left( 5-3 \right)!} \\
 & \Rightarrow P_{3}^{5}=\dfrac{5!}{2!} \\
 & \Rightarrow P_{3}^{5}=\dfrac{5\times 4\times 3\times 2!}{2!} \\
 & \Rightarrow P_{3}^{5}=\dfrac{5\times 4\times 3\times 1}{1} \\
 & \Rightarrow P_{3}^{5}=20\times 3 \\
 & \Rightarrow P_{3}^{5}=60 \\
\end{align}$
Now the other two places can be filled by $P_{3}^{5}=60$ ways. Therefore, the total number of possibilities here is given by
$\begin{align}
  & 1\times P_{3}^{5}=1\times 60 \\
 & \Rightarrow 1\times P_{3}^{5}=60 \\
\end{align}$
Hence the total number of possibilities is 60.

Note: Alternatively we can use a filling up method to fill the rest of three places. This is done below.
We can form any number by 2, 3, 4, 5, 6 and 7 digits as 2 has been fixed. In order to form a four digit number first digit should be free from 0. But here we are not given 0. Therefore now we have fixed 2 there. Therefore, for the first place we have one choice only. So we now have $\underline{1}\,\underline{{}}\,\underline{{}}\,\underline{{}}$. We can consider the other three places can be filled by any number from 3, 4, 5, 6 and 7 digits. But in this question the limitation given to us is that the repetition is not allowed. Therefore, as we have total digits 2, 3, 4, 5, 6 and 7 and out of these one is already used, fixed. Therefore, we are left with now 5 choices. That is $\underline{1}\,\underline{5}\,\underline{{}}\,\underline{{}}$. And the next place has 4 choices as there is no repetition. That is $\underline{1}\,\underline{5}\,\underline{4}\,\underline{{}}$. And at last we have 3 choices thus we get $\underline{1}\,\underline{5}\,\underline{4}\,\underline{3}$. Therefore, by multiplying the choices we have a total number of possibilities given by $1\times 5\times 4\times 3=60$.
Hence the total number of possibilities is 60.