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 =

= \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.