Question

How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7 if repetition of digits is not allowed?

(a)\[{}^{7}{{P}_{3}}-{}^{6}{{P}_{3}}\]
(b)\[{}^{10}{{P}_{3}}-{}^{6}{{P}_{2}}\]
(c)\[{}^{7}{{P}_{3}}-{}^{6}{{P}_{2}}\]
(d)\[{}^{8}{{P}_{3}}-{}^{7}{{P}_{2}}\]

Answer 1

Hint: The numbers between 100 to 1000 will be 3 digit numbers. The required numbers will be a total 3 digit number possible minus the number of digits that can start with zero.

Complete step-by-step answer:

We need to find the number of digits that can be formed between 100 and 1000 with the digits 0, 1, 2, 3, 4, 5, 6, 7. The number between 100 and 1000 will be a 3 digit number.

The 3 digit number will be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7.

But these include numbers starting with zero like 025, 035 etc, which are actually 2 digit numbers.

\[\therefore \] Required numbers = Total 3 digit numbers – 3 digit number which has zero in the beginning.

First let us find total 3 digit numbers.

Total digits in 0, 1, 2, 3, 4, 5, 6, 7 = 8

\[\therefore \] n = 8

Here, we have to form 3 – digit numbers, so let’s take r = 3.

\[\therefore \] Total number of 3 digits number = \[{}^{n}{{P}_{r}}={}^{8}{{P}_{3}}\]

\[\begin{align}

  & {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} \\

 & \therefore {}^{8}{{P}_{3}}=\dfrac{8!}{\left( 8-3 \right)!}=\dfrac{8!}{5!}=\dfrac{8\times 7\times 6\times 5!}{5!} \\

 & \therefore {}^{8}{{P}_{3}}=8\times 7\times 6=336 \\

\end{align}\]

\[\therefore \] Total number of 3 digit number \[={}^{8}{{P}_{3}}=336\]

Now let us find the 3 digit numbers starting with zero.

Total digits left = 1, 2, 3, 4, 5, 6, 7

\[\therefore \] n = 7

Number of digits to be filled = 2. So, let’s take, r = 2.

\[\therefore \] Total number of 3 digits numbers starting with zero \[={}^{n}{{P}_{r}}={}^{7}{{P}_{2}}\]

                                                                                                   \[\begin{align}

  & =\dfrac{7!}{\left( 7-2 \right)!}=\dfrac{7!}{5!} \\

 & =\dfrac{7\times 6\times 5!}{5!} \\

 & =7\times 6=42 \\

\end{align}\]

Hence the total number of 3 digit numbers starting with zero = 42

\[\therefore \] Required number = Total 3 digit numbers – 3 digit number which has zero in starting

                                      = 336 – 42 = 294 = \[{}^{8}{{P}_{3}}-{}^{7}{{P}_{2}}\]

Hence the total number lying between 100 and 1000 which can be formed with digits 0, 1, 2, 3, 4, 5, 6, 7 are 294 numbers i.e. \[{}^{8}{{P}_{3}}-{}^{7}{{P}_{2}}\].

\[\therefore \] Option (d) is the correct answer.


Note: In a question like this be careful as to subtract 3 digit numbers starting with zero. If zero was not given but only 1, 2, 3, 4, 5, 6, and 7 were given as digits, then the required number will be equal to the 3 digit numbers formed.