Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 5"
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==Problem== | ==Problem== | ||
| − | How many 10-digit positive | + | How many 10-[[digit]] [[positive integer]]s have all digits either 1 or 2, and have two consecutive 1's? |
==Solution== | ==Solution== | ||
| + | We take as our universe the set of 10-digit [[integer]]s whose digits are all either 1 or 2, of which there are <math>2^{10}</math>, and we count the complement. The complement is the set of 10-digit positive integers composed of the digits 1 and 2 with no two consecutive 1s. Counting such numbers is a popular combinatorial problem: we approach it via a [[recursion]]. | ||
| − | + | There are two "good" one-digit numbers (1 and 2) and three good two-digit numbers (12, 21 and 22). Each such <math>n</math>-digit number is formed either by gluing "2" on to the end of a good <math>(n - 1)</math>-digit number or by gluing "21" onto the end of a good <math>(n - 2)</math>-digit number. This is a [[bijection]] between the good <math>n</math>-digit numbers and the [[union]] of the good <math>(n-1)</math>- and <math>(n - 2)</math>-digit numbers. Thus, the number of good <math>n</math>-digit numbers is the sum of the number of good <math>(n-1)</math>- and <math>(n - 2)</math>-digit numbers. The resulting recursion is exactly that of the Fibonacci numbers with initial values <math>F_1 = 2</math> and <math>F_2 = 3</math>. | |
| + | Thus, our final answer is <math>2^{10} - F_{10} = 1024 - 144 = 880</math>. | ||
| − | ---- | + | ==Solution2== |
| + | Again, we seek the amount without two consecutive 1s. Assume there are n 2s in one of the numbers. We have to insert the 10-n 1s in the n+1 slots surrounding the 2s. Thus, there are <math>{{n+1} \choose{10-n}}</math> possibilities for a number with n 2s. The answer is thus <math>2^{10}- {6 \choose 5} - {7 \choose 4} - {8 \choose 3} - {9 \choose 2} - {10 \choose 1} - {11 \choose 0}=880</math>. | ||
| − | + | == See also == | |
| − | + | {{Mock AIME box|year=2006-2007|n=4|num-b=4|num-a=6|source=125025}} | |
| − | + | ||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
Latest revision as of 13:50, 16 February 2009
Contents
Problem
How many 10-digit positive integers have all digits either 1 or 2, and have two consecutive 1's?
Solution
We take as our universe the set of 10-digit integers whose digits are all either 1 or 2, of which there are 
, and we count the complement. The complement is the set of 10-digit positive integers composed of the digits 1 and 2 with no two consecutive 1s. Counting such numbers is a popular combinatorial problem: we approach it via a recursion.
There are two "good" one-digit numbers (1 and 2) and three good two-digit numbers (12, 21 and 22). Each such 
-digit number is formed either by gluing "2" on to the end of a good 
-digit number or by gluing "21" onto the end of a good 
-digit number. This is a bijection between the good -digit numbers and the union of the good 
- and -digit numbers. Thus, the number of good -digit numbers is the sum of the number of good - and -digit numbers. The resulting recursion is exactly that of the Fibonacci numbers with initial values 
and 
.
Thus, our final answer is 
.
Solution2
Again, we seek the amount without two consecutive 1s. Assume there are n 2s in one of the numbers. We have to insert the 10-n 1s in the n+1 slots surrounding the 2s. Thus, there are 
possibilities for a number with n 2s. The answer is thus 
.
See also
| Mock AIME 4 2006-2007 (Problems, Source) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
