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Difficult Math Problem #60 - Combinations

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Re: Difficult Math Problem #60 - Combinations

by kulksnikhil » Thu Nov 23, 2006 4:57 pm
800guy wrote:If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there?

A. 42,500
B. 37,500
C. 45,000
D. 40,000
E. 50,000
Last digit can have 5 numbers (1,3,5,7,9)
Fourth to Second digits can have 10 numbers
First digit can have 8 out of 10 digitis (leaving 0 and 5)

So number of possibilities:

8 * 10 * 10 * 10 * 5 = 40,000

Answer: D
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by gmat_enthus » Fri Nov 24, 2006 7:28 am
Agree D
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OA

by 800guy » Fri Nov 24, 2006 11:23 am
OA:

This problem can be solved with the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

There are 8 possibilities for the first digit (1, 2, 3, 4, 6, 7, 8, 9).
There are 10 possibilities for the second digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the third digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the fourth digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 5 possibilities for the fifth digit (1, 3, 5, 7, 9)

Using the Multiplication Principle:

= 8 * 10 * 10 * 10 * 5
= 40,000
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my approach!!

by Rashmi1804 » Tue Mar 03, 2009 10:05 am
i got a totally different answer...my answer is 4032 ways!!

please go through...

CASE-1
number of ways in which first place can be filled 8c1 ( since 0 and 5 are eliminated)

number of ways in which fifth place can be filled [when first place has an ODDnumber] = 4c1 ( only four of " 1,3,5,7,9" , because one of them is already used above)

number of ways in which the middle three places can be filled = 8c3 ( since two of the 10 digits are already taken by first and fifth places)

=8*4*(8!/3!*5!)

CASE-2
number of ways in which first place can be filled= 8c1 (not 10 since 0 and 5 are eliminated)

number of ways in which fifth place can be filled [when first place has an EVEN number] = 5c1 (1,3,5,7,9)

number of ways in which the middle three places can be filled = 8c3 ( since two of the 10 digits are already taken by first and fifth places)

=8*5*(8!/3!*5!)


this gives= 4032 ways!!


but none of the options is 4032.

there must be blunder i am comitting in this solution or the answer choices are wrong.

Could some expert please look into this ??
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by Rashmi1804 » Tue Mar 03, 2009 10:13 am
okiee!! i jus got a clue after dwelling into the problem for some more time...IS IT BECAUSE A DIGIT CAN BE REPEATED MY APPROACH IS WRONG ??
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by quocbao » Tue Mar 03, 2009 7:54 pm
Yes, each the of the digit can be repeated. They don't say anything about a number can only appear one.
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Answer

by Dream Weaver » Tue Mar 03, 2009 9:36 pm
Agreed D.

8 * 10 * 10 * 10 * 5 =40,000
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Answer

by Dream Weaver » Tue Mar 03, 2009 9:37 pm
Agreed D.

8 * 10 * 10 * 10 * 5 =40,000
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