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Question: How many numbers lying between \[999\] and \[10000\] can be formed with the help of the digits \[\;0...

How many numbers lying between 999999 and 1000010000 can be formed with the help of the digits   0,2,3,6,7,8\;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

Explanation

Solution

We have to find the total numbers lying between 999999 and 1000010000 which can be formed with the help of the digits   0,2,3,6,7,8\;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 00 as if the number starts with 00 then it becomes a three digit number and doesn't lie between the numbers 999999 and 1000010000 .

Complete step-by-step solution:
Given :
The numbers formed should lie between 999999 and 1000010000 , so a four digit number should be formed using the given digits .
Now ,
The digits which can be placed on the thousands place are :- 2,3,6,72,3,6,7 and 88
( 00 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 = 5
Now , one of these digits is placed at thousands places so we are left with a total 55 more digits ( including 00 ) 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= 5
Now , one more of these digits is placed at hundreds place so we are left with a total 44 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 =4
Now , one more of these digits is placed at tens place so we are left with a total 33 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 = {}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 =5×5×4×3 = 5\times 5\times 4\times 3
Total number which can be formed using the digits =300 = 300
Thus , the total numbers between 999999 and 1000010000 are 300300 .
Hence , the correct option is (3)\left( 3 \right)

Note: We can also directly solve this question using the formula of combination I.e. total number of digits =5×3!×5C3 = 5\times 3!\times {}^5{C_3}
corresponding to each combination of nCr{}^n{C_r} we have r!r! permutations, because rr objects in every combination can be rearranged in r!r!ways . Hence , the total number of permutations of n different things taken rr at a time isnCr×r!{}^n{C_r} \times {}r!. Thus   nPr=nCr×r!,  0<r⩽n\;{}^n{P_r}{} = {}^n{C_r} \times r!,\;0 < r \leqslant n
Also , some formulas used :
nC1=n{}^n{C_1} = n
nC2=n(n−1)2{}^n{C_2}{} = \dfrac{{n\left( {n - 1} \right)}}{2}
nC0=1{}^n{C_0}{} = 1
nCn=1{}^n{C_n} = 1