Another Counting Problem

[knight247]

How many 5-digit positive integers exist where no two consecutive digits are the
same?
A) 9*9*8*7*6
B) 9*9*8*8*8
C) 9^5
D) 9*8^4
E) 10*9^4

OA is C

[shankar.ashwin]

I think slot method would work well here;

_ _ _ _ _

1st - (1-9) - 9 possibilities
2nd - (10 - 1) 9 poss (Excluding the 1st digit)
3rd - (10-1) 9 possibilities (Excluding 2nd digit)
and so on.

You get 9^5 C

[GmatKiss]

I think slot method would work well here;

_ _ _ _ _

1st - (1-9) - 9 possibilities
2nd - (10 - 1) 9 poss (Excluding the 1st digit)
3rd - (10-1) 9 possibilities (Excluding 2nd digit)
and so on.

You get 9^5 C

Hi Shankar,

I am unable to follow.
Could you please explain in detail.

TIA,
GK

[zooki]

say you have following Slots for the 5 digits

_ _ _ _ _

for slot 1 you have 9 possibilities from 1 -9 (excluding 0, because that make it 4 digit #)
for slot 2 you have 9 possibilities from 1- 0 (excluding the previous one)
and so on

Therefore, 9 possibilities for each slot. 9^5

[sam2304]

I agree with shankar. IMO C

_ _ _ _ _
9 9 9 9 9

Condition is no two consecutive integers should be same but it can occur again as the 3rd digit. So every place has 9 choices excluding the digit which occurred previously. Nice problem

[shankar.ashwin]

First thing to note is this problem allows repetition, so all digits could take 10 values (from 0-9) except the first (which cannot take 0)

If there were no restrictions we would have 9*10*10*10*10 5 digit integers possible.

But we have a condition where no 2 digits can be consecutive. (Say I pick the 1st digit as '1', then the 2nd can take any value except '2') Thats 9 options.

So, 1st position has 9 possibilities,
2nd has 9,
3rd again has 9 (Now if the 2nd is say '5', 3rd can take any value expect '6')
and so on.

Together, we get 9^5.

Hope you understood

Hi Shankar,

I am unable to follow.
Could you please explain in detail.

TIA,
GK