Question

How many numbers less than $10,000$ can be made from \[1,2,3,4,5,6,7,0\]?
A). $1820$
B). $1821$
C). $1900$
D). $1901$

Answer 1

Hint- In order to deal with this question we will use permutation and combination here will will make the case for \[1\] digit, \[2\] digit , \[3\]digit and \[4\] digits number and we will apply permutation to find the total numbers in each case.

Complete step-by-step solution -
Given digits are \[1,2,3,4,5,6,7,0\]
We have to form number which is less than $10,000$
Largest number of \[4\] digit is \[\;9,999\]
Now we will calculate total number formed by one digit, two digit, three digits and by \[4\] digits
For one digit number:
There are $8$ possibilities to make one digit number which is less than $10,000$ they are \[0,1,2,3,4,5,6,7,8{\text{ }} = {\text{ }}8\]
So the total number of one digit number \[ = 8\]
For two digit number:
To find the \[2\] digit number first we have to find total number formed by \[2\] digits further we have to subtract all the number which is started from \[0\]
As we know that if we have to choose \[r\] number from total \[n\] number and shuffling of chosen number will also allowed than it is presented as
\[{}^n{C_r} \times r!\]
By using the above property
Total number of \[2\] digits \[ = {}^8{C_2} \times 2!\]
And the total numbers which are started from \[0{\text{ }} = {\text{ }}7{\text{ }}\left( {{\text{ }}01,02,03,04,05,06,07} \right)\]
Therefore required two digit numbers =
\[ {}^8{C_2} \times 2! - 7 \\
   = \dfrac{{8!}}{{6!2!}} \times 2! - 7 \\
   = \dfrac{{8!}}{{6!}} - 7 \\
   = 56 - 7 \\
   = 49 \\ \]
For three digit number:
Similarly , To find the \[3\] digit number first we have to find total number formed by \[3\] digits further we have to subtract all the number which is started from \[0\]
As we know that if we have to choose \[r\] number from total \[n\] number and shuffling of chosen number will also allowed than it is presented as
\[{}^n{C_r} \times r!\]
By using the above property
Total number of \[3\] digits \[ = {}^8{C_3} \times 3!\]
And the total numbers which are started from \[0 = {}^7{C_2} \times 2!\]
Therefore required three digit numbers =
\[ {}^8{C_3} \times 3! - {}^7{C_2} \times 2! \\
   = \dfrac{{8!}}{{5!3!}} \times 3! - \dfrac{{7!}}{{5!2!}} \times 2! \\
   = \dfrac{{8!}}{{5!}} - \dfrac{{7!}}{{5!}} \\
   = \dfrac{{8! - 7!}}{{5!}} \\
   = 294 \\ \]
For four digit number:
Again to find the \[4\] digit number first we have to find total number formed by \[4\] digits further we have to subtract all the number which is started from \[0\]
As we know that if we have to choose \[r\] number from total \[n\] number and shuffling of chosen number will also allowed than it is presented as
\[{}^n{C_r} \times r!\]
By using the above property
Total number of \[4\] digits \[ = {}^8{C_4} \times 4!\]
And the total numbers which are started from \[0 = {}^7{C_3} \times 3!\]
Therefore required three digit numbers =
\[ = {}^8{C_4} \times 4! - {}^7{C_3} \times 3! \\
   = \dfrac{{8!}}{{4!4!}} \times 4! - \dfrac{{7!}}{{4!3!}} \times 3! \\
   = \dfrac{{8!}}{{4!}} - \dfrac{{7!}}{{4!}} \\
   = \dfrac{{8! - 7!}}{{4!}} \\
   = 1470 \\ \]
Total numbers formed by given digits = One digit numbers + two digits number + three digits number + four digits number
\[ = 8 + 49 + 294 + 1470 \\
   = 1821 \\ \]
Hence the correct answer is option B.

Note- The combination or shorter \[{}^n{C_r}\] is the number of ways we can pick \[r\] objects from a set containing \[n\] different objects, so that (unlike permutations) the selection order doesn't matter. The \[C{\text{ }}\left( {n,{\text{ }}r} \right)\]symbol refers to the number of combinations of \[n\] items taken \[r\] at a time.