Question

How many numbers lying between \[999\] and \[10000\] can be formed with the help of the digits \[\;0,2,3,6,7,8\] when the digits are not repeated?
A). \[\left( 1 \right)\]\[100\]
B). \[\left( 2 \right)\]\[200\]
C). \[\left( 3 \right)\]\[300\]
D). \[\left( 4 \right)\]\[400\]

Answer 1

Hint: We have to find the total numbers lying between \[999\] and \[10000\] which can be formed with the help of the digits \[\;0,2,3,6,7,8\] when the digits are not repeated . We solve the question using the concept of permutation and combination . a four digit number is to be formed using the given digits . Also , to consider that a number cannot start with $0$ as if the number starts with $0$ then it becomes a three digit number and doesn't lie between the numbers \[999\] and \[10000\] .

Complete step-by-step solution:
Given :
The numbers formed should lie between \[999\] and \[10000\] , so a four digit number should be formed using the given digits .
Now ,
The digits which can be placed on the thousands place are :- \[2,3,6,7\] and $8$
( \[0\] cannot be placed at thousands place as putting it at thousands place makes the number a three digit number . )
So , the number of ways in which thousands place can be filled \[ = 5\]
Now , one of these digits is placed at thousands places so we are left with a total \[5\] more digits ( including $0$ ) as repetition of digits is not allowed .
Now , any of the left digits can be placed at hundreds place
So , the number of ways in which hundreds place can be filled \[= 5\]
Now , one more of these digits is placed at hundreds place so we are left with a total $4$ more digits as repetition of digits is not allowed .
Now , any of the left digits can be placed at tens place
So , the number of ways in which tens place can be filled \[ =4\]
Now , one more of these digits is placed at tens place so we are left with a total $3$ more digits as repetition of digits is not allowed .
Now , any of the left digits can be placed at ones place
So , the number of ways in which ones place can be filled \[ = {}3\]
Total number which can be formed using the digits = number of digit at thousands place $\times$ number of digit at hundreds place $\times$ number of digit at tens place $\times$ number of digit at ones place
Total number which can be formed using the digits \[ = 5\times 5\times 4\times 3\]
Total number which can be formed using the digits \[ = 300\]
Thus , the total numbers between \[999\] and \[10000\] are \[300\] .
Hence , the correct option is \[\left( 3 \right)\]

Note: We can also directly solve this question using the formula of combination I.e. total number of digits \[ = 5\times 3!\times {}^5{C_3}\]
corresponding to each combination of ${}^n{C_r}$ we have \[r!\] permutations, because $r$ objects in every combination can be rearranged in \[r!\]ways . Hence , the total number of permutations of n different things taken $r$ at a time is\[{}^n{C_r} \times {}r!\]. Thus \[\;{}^n{P_r}{} = {}^n{C_r} \times r!,\;0 < r \leqslant n\]
Also , some formulas used :
\[{}^n{C_1} = n\]
\[{}^n{C_2}{} = \dfrac{{n\left( {n - 1} \right)}}{2}\]
\[{}^n{C_0}{} = 1\]
\[{}^n{C_n} = 1\]