Counting Calamity

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tabsang: Senior | Next Rank: 100 Posts; Posts: 64; Joined: Mon Jun 13, 2011 5:27 am; Thanked: 5 times

Counting Calamity

by tabsang » Sun Dec 16, 2012 4:31 am

How many 4-digit numbers can be formed by using the digits 0-9, so that the numbers contains exactly 3 distinct digits?

(A) 1944
(B) 3240
(C) 3850
(D) 3888
(E) 4216

I got [spoiler](D)[/spoiler] in a little over 3.5 minutes and I don't even know if it's right :O

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Source: Beat The GMAT — Problem Solving | Sun Dec 16, 2012 4:31 am

puneetkhurana2000: Master | Next Rank: 500 Posts; Posts: 131; Joined: Wed Nov 14, 2012 2:01 pm; Thanked: 39 times; Followed by:2 members

by puneetkhurana2000 » Sun Dec 16, 2012 2:51 pm

It should be 9*9*8 * 4C2 = 3888.

First 9 for the '000 digit as 0 can't come.
Second 9 for the '00 digit as 0 can come but not the digit that has already been at the '000 place.
Third 8 for the '0 digit. Same logic as above.
4C2 as any of the 3 digits can be repeated but order of digits does matter, so 4*3/2 = 6

Answer D.

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tabsang: Senior | Next Rank: 100 Posts; Posts: 64; Joined: Mon Jun 13, 2011 5:27 am; Thanked: 5 times

by tabsang » Sun Dec 16, 2012 10:51 pm

Well, here's my approach:

Case I:
The repeated digit is the unit's digit.
So, the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways, respectively.
Now the 4th digit (unit's digit) can be either equal to the 1st, 2nd or 3rd digit.
Thus, in all we have:
9x9x8x3

Case II:
The repeated digit is the ten's digit.
So, the 1st, 2nd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the 3rd digit (ten's digit) can be either equal to the 1st or 2nd digit.
Thus, in all we have:
9x9x2x8

Case III:
The repeated digit is the hundred's digit.
So, the 1st, 3rd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the 2nd digit (hundred's digit) is equal to the 1st digit.
Thus, in all we have:
9x1x9x8

In totality, we have 9x9x8(3+2+1) = 9x9x8x6 = [spoiler](D) 3888[/spoiler]

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tabsang: Senior | Next Rank: 100 Posts; Posts: 64; Joined: Mon Jun 13, 2011 5:27 am; Thanked: 5 times

by tabsang » Sun Dec 16, 2012 10:59 pm

puneetkhurana2000 wrote:It should be 9*9*8 * 4C2 = 3888.

First 9 for the '000 digit as 0 can't come.
Second 9 for the '00 digit as 0 can come but not the digit that has already been at the '000 place.
Third 8 for the '0 digit. Same logic as above.
4C2 as any of the 3 digits can be repeated but order of digits does matter, so 4*3/2 = 6

Answer D.

Hey Puneet,

Could you explain the last part of your solution in more detail?
I didn't follow it.
If you say that the order of digits matters then use of permutations is in order right?
But you've used combinations (4C2).
Moreover, you've used 4C2... i.e (4 options for two slots????).
Now, we only have one slot and that can be filled by a digit that is same as one of the other 3 digits in the number, right?? And then we get an arrangement with exactly 3 distinct digits.

TIA

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

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