Another Counting Problem

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Sat Oct 15, 2011 12:21 am

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

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Source: Beat The GMAT — Problem Solving | Sat Oct 15, 2011 12:21 am

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Sat Oct 15, 2011 12:43 am

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

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GmatKiss: Legendary Member; Posts: 2789; Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members; Location: Chennai, India; Thanked: 206 times; GMAT Score:640

by GmatKiss » Sat Oct 15, 2011 3:21 am

shankar.ashwin wrote: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

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zooki: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Fri Oct 14, 2011 10:15 amFollowed by: 1 member; Thanked: 4 times

by zooki » Sat Oct 15, 2011 5:44 am

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

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sam2304: Legendary Member; Posts: 1239; Joined: Tue Apr 26, 2011 6:25 amFollowed by: 26 members; Thanked: 233 times; GMAT Score:680

by sam2304 » Sat Oct 15, 2011 5:52 am

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

Getting defeated is just a temporary notion, giving it up is what makes it permanent.
https://gmatandbeyond.blogspot.in/

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Sat Oct 15, 2011 7:57 am

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

GmatKiss wrote:
Hi Shankar,

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

TIA,
GK

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