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Back Again!!! P&C

by sudhir3127 » Sat Aug 16, 2008 9:33 pm
1. a)

X has forgotten the telephone number of his best friend Y. All he remembers is that the number has 8 digits and ended with an odd number and had exactly one 9. How many possible ways does X have to try to be sure that he gets the correct number.

b) If in the above probelm X is reminded by his another friend Z that apart from what he remembered there was the additional fact that the last digit of the number was not repeated under any circumstance, then how many ways does X have to try to be sure that he gets the correct number.

Answers after some discussions.....
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Re: Back Again!!! P&C

by Stuart@KaplanGMAT » Sat Aug 16, 2008 9:36 pm
sudhir3127 wrote:1. a)

X has forgotten the telephone number of his best friend Y. All he remembers is that the number has 8 digits and ended with an odd number and had exactly one 9. How many possible ways does X have to try to be sure that he gets the correct number.

b) If in the above probelm X is reminded by his another friend Z that apart from what he remembered there was the additional fact that the last digit of the number was not repeated under any circumstance, then how many ways does X have to try to be sure that he gets the correct number.

Answers after some discussions.....
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by pepeprepa » Sun Aug 17, 2008 1:51 am
Stuart the thing is that the two questions are totally related but it would be nice to check oneself with answer choices in order not to make silly mistakes.


1)
The question is how many different phone numbers of this style exist?

The number 9 has 7 positions in the phone number given the last one place is for an even number.
The last digit of the phone number is an even number, 0, 2, 4, 6, 8 --> 5 possibilities.
For the other remaining 6 places we can choose any number except 9 --> 9 possibilities (0 1 2 3 4 5 6 7 8 )

7*5*9^6


2)
The difference is that we cannot use the last digit (the even one) at the other places of the phone number.
We have to change the possibilities for the remaining 6 places --> We now have 8 possibilities ( 0 1 2 3 4 5 6 7 8 - the last even number)

7*5*8^6
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by sudhir3127 » Sun Aug 17, 2008 2:31 am
the answer choices for
a are

1.104*9^5
2.113*9^5
3.300*9^5
4.764*9^5*6!

for b are

1. 200*8^5+ 72*9^5
2.8*96
3.36*85 + 7*96
4.36*8^5 + 8*9^6

hope this helps..

pepeprepa.. i think u made a mistake in considering the last digits as even .. the question says it shud be odd...
Last edited by sudhir3127 on Sun Aug 17, 2008 5:28 am, edited 1 time in total.
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by pepeprepa » Sun Aug 17, 2008 3:17 am
Yep fool error.

For the 1)
There are 8 places for the digit 9.

When 9 is the last one digit of the phone number: _ _ _ _ _ _ _ 9
9^7

When 9 is at any other place (7 places):
I multiply by 7 for the 7 different places of 9.
I multiply by 4 for the different possibilities of the last digit of the phone number which is odd ( 1 3 5 7, 9 is not included given he must not repeat)
The remaining places have 9 possibilitties each:
7*4*9^6

I arrive at 9^7 + 7*4*9^6= 333*9^5

You see where I am wrong?
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by sudhir3127 » Sun Aug 17, 2008 3:20 am
the OA are 3 and 1. i am not sure how to get.. even i am finding it difficult to get the answers....
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by pepeprepa » Sun Aug 17, 2008 4:53 am
Can you describe what you do for 1) ?
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by 4meonly » Sun Aug 17, 2008 4:58 am
sudhir3127 wrote:the OA are 3 and 1. i am not sure how to get.. even i am finding it difficult to get the answers....
?Really?

I've got

5*7*9^6

and 4*7*8^6
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by sudhir3127 » Sun Aug 17, 2008 5:26 am
pepeprepa wrote:Can you describe what you do for 1) ?
it took me a lot of time to solve this (1/2 hr ) but i did only the first part..

here it goes..

there are 3 situations in this ...
1.when 9 is in the units digit ..
2. when 9 is first digit
3. when 9 is neither the last nor the first digit.

situation 1.

when 9 is in the units digit.

9 can arrange itself in last place in 1 way.
(0-8 ) digits can arrange themselves in 1 st place in 8 ways ( because we have 0 in this)
rest 6 places we can arrange in 9^6 ways..

thus we have 9^6*8*1= 72*9^5

situation 2.
when 9 is first digit

9 can arrange itself in the first place in 1 way..
last digit has to be an odd number from 1,3,5,7. in 4 ways..
remaining can arrange themselves in 9^6 ways
thus we have 1*4*9^5= 36*9^5 ways

situation 3.
when 9 is neither the last nor the first digit
last digit shud be arranged from 1,3,5,7 in 4 ways
first digit in 8 ways
9 can arrange itself in 6 places in 6 ways,
rest 5 can arrange themselves in 9^5 ways.
thus we have 4*8*6*9^5= 192*9^5 ways..

there fore the total is
72*9^5 + 36*9^5 + 192*9^5 = 300*9^5 ways,...

Hope its clear... please let me know if u have any doubts...
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by sudhir3127 » Sun Aug 17, 2008 5:41 am
sudhir3127 wrote:the answer choices for
a are

1.104*9^5
2.113*9^5
3.300*9^5
4.764*9^5*6!

for b are

1. 200*8^5+ 72*9^5
2.8*96
3.36*85 + 7*96
4.36*8^5 + 8*9^6

hope this helps..

pepeprepa.. i think u made a mistake in considering the last digits as even .. the question says it shud be odd...
this is how we do the second bit...

taking the same condition + the last digit cant repeated again.

situation 1.

when 9 is in the units digit.

9 can arrange itself in last place in 1 way.
first digit can arrange itself in 8 ways.
remaining 6 places can arrange themselves in 9^6 ways
total = 1*8*9^6 = 72*9^5 ways

situation 2.
when 9 is first digit

first place in 1 way.
last place in 4 ways from (1,3,5,7)
rest 6 places out of 8 in 8^6 ways
total = 1*4*8^6 ways = 32*8^5 ways

situation 3.
when 9 is neither the last nor the first digit

lasr place in 4 ways ( from 1,3,5,7 )
first place in 7 ways ( 9 cant take it )
9 can arrange itself in 6 places in 6 ways
rest 5 places out of 8 in 8^5 ways..
total 4*7*6*8^5 ways = 168*8^5 ways

thus the total is
72*9^5 + 32*8^5 + 168*8^5=
200*8^5 + 72*9^5..

hope it helps..

do let me know if u have any doubts...
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by Ian Stewart » Sun Aug 17, 2008 7:19 am
Pepeprepa's solution (after fixing the even/odd error) looks perfect to me, and that's exactly how I'd approach the question as well. Of course, it doesn't lead to one of the four answer choices, and that seems to be because we're supposed to assume a phone number cannot begin with zero. But how are we supposed to guess this? In the UK, every mobile phone number begins with zero, so telephone number formatting is different depending where you live. The GMAT doesn't test whether you know how telephone numbers are formatted- if we are supposed to assume the phone number cannot begin with zero, a real GMAT question would always mention this. So, I'm curious about the source of the problem.

In any case, it's the type of problem you would need to answer by doing a case-by-case analysis, as Sudhir and Pepeprepa have done.
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