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Counting numbers with sum = 2

Expert replies
by Brent@GMATPrepNow » Thu Jan 01, 2009 8:04 am
How many integers between 1 and 10^81 are such that the sum of their digits is 2?
A) 3160
B) 3240
C) 3241
D) 3320
E) 3321
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Source: — Problem Solving |

by daisy » Thu Jan 01, 2009 9:31 am
IMO (E)

Sum of all the digits will be 2 only when either have only if one appears twice in a number and rest should be zero or there is 1 two present in the number and rest should be zero.

for example,
Up to 4 digits:-

Case 1:
1100
1001
1010
0011

Case 2:
2000
0200
0020
0002

Since the upper limit is 10^81
therefore there will 81 places where we can arrange the combination of 2 ones and zeros or 1 two and zeros.

For case 1:
The total number of cases will be:
81C1 x 80C80 = 81

For case 2:
81C2 x 79C79 = 3240

Therefore total cases: 3240 + 81 = 3321

Choose (E)
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by Brent@GMATPrepNow » Thu Jan 01, 2009 10:36 am
Nice work, Daisy. The answer is E

I'll post my solution as well:
Begin by noting that, if the sum is 2, then there are two cases:
Case 1: The number has two 1’s and all remaining digits are 0
Case 2: The number has one 2 and all remaining digits are 0
If a number is less than 10^81, then that number has 81 digits or fewer. For this question, we’ll say that all numbers less than 10^81 have EXACTLY 81 digits, however some of those numbers have leading digits of zero (e.g., 0000…0001001 = 1001 and 0000…00200 = 200).
Case 1: In how many ways can we have an 81-digit number such that there are two 1’s and seventy-nine 0’s? Essentially, we have 81 places and we must select 2 of those places to hold the digit “1.” We can accomplish this selection in 81C2 (3240) ways.
So, there are 3240 numbers consisting of two 1’s

Case 2: In how many ways can we have an 81-digit number such that there is one 2 and the remaining digits are 0? We have 81 places and we must select 1 place to hold the 2 (the remaining places will be filled with 0’s). There are 81 places and we must select 1 place. We can accomplish this in 81 ways.
Adding the results from Cases 1 and 2, we get a total of 3321 numbers.
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by logitech » Thu Jan 01, 2009 10:44 am
Brent, I am a big fan of your questions and this one is one of my favorites. Great job!
LGTCH
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