Question

The number of six digit positive even numbers in which no two consecutive digits are same will be

• A. ${5.9}^5$
• B. ${}^5{C}_3\times \frac{9^5}{2}$
• C. $\frac{1}{2}(9^6+1)$
• D. $\frac{1}{2}(9^6-1)$

Answer 1

The correct option is C $\frac{1}{2}(9^6+1)$

The unit's place of the six digit number can be filled in $5$ ways.

Then, the place next to it can be filled in $9$ ways. (since no consecutive digits are same). Similarly the remaining places can also be filled in $9$ ways each.

Hence, total number of ways of filling $6$ places such that unit's place is an even number and no two consecutive digits are same is

$9\times 9\times 9\times 9\times 9=9^5$

But, this will include cases where the left most place is $0$. In that case we shall get $5$ digit numbers satisfying the condition that unit's place is even and no two consecutive digits are same.

Hence, if we denote the number of such $6$ digit numbers possible as $N_6$ and $5$ digit numbers possible as $N_5$, then

$N_6+N_5=5\times 9^5$...(1)

Similarly, $N_5+N_4=5\times 9^4$...(2)

$N_4+N_3=5\times 9^3$...(3)

$N_3+N_2=5\times 9^2$...(4)

$N_2+N_1=5\times 9$...(5)

(1) - (2) + (3) - (4) + (5) gives $N_6+N_1=5(9^5-9^4+9^3-9^2+9)=5\left[\frac{9(1-(-9{)}^5)}{1-(-9)}\right]=\frac{9^6+9}{2}$

$N_1$ is the number of one digit positive even number which is $4$.

Hence, $N_6=\frac{9^6+9}{2}-4$

$=\frac{9^6+1}{2}$