#### The number of 5-digit integers formed by the digits 0,3,4,5,6,8 in the interval [3502, 5472] isOption: 1 4855Option: 2 3955Option: 3 5455  Option: 4 6595

Given that,

The number of 5-digit integers formed by the digits 0,3,4,5,6,8 in the interval [3502, 5472].

Let us consider the 1st digit as A, the 2nd digit as B, the 3rd digit as C, the 4th digit as D, and the 5th digit as E.

Position A has the digits 3, 4, and 5.

If the digits 0, 6, and 8 are put in position A then it will not be in the range [3502, 5472].

So, position A can be filled with 3, 4, and 5 in $^{3}C_{1}=3$$^{3}C_{1}=3$ ways.

Position B can be filled with any one of the digits 0, 3, 4, 5, 6, and 8 in $^{6}C_{1}=6$ ways.

Position C can be filled with any one of the digits 0, 3, 4, 5, 6, and 8 in $^{6}C_{1}=6$ ways

Position D can be filled with any one of the digits 0, 3, 4, 5, 6, and 8 in $^{6}C_{1}=6$ ways.

Position E can be filled with any one of the digits 0, 3, 4, 5, 6, and 8 in $^{6}C_{1}=6$ ways.

Thus, the number of ways of the number starting with 3, 4, and 5 is given by,

$3\times 6\times 6\times 6\times 6=3888$

Now, we have to find the number which starts with 3 and 5 but doesn’t fall in the range [3502, 5472].

The possible way of starting the number with 300 is 6.

Placing 0 in the last place in 3502 will be less than the range.

So, the possible way is 1.

Considering the number 5472, here the 2nd place is filled with 5 or 6 will go beyond the range. In the 3rd place, any of the 6 digits can be placed.

So, the possible way is $2\times 5\times 6\times =60$.

The total number which starts with 3 and 5 but doesn’t fall in the range is,

$6+1+60=67$

Therefore, the total number falls in the range is 3888+67 = 3955.