Sequencing the Digits
How many ways are there to write the numbers 0 through 9 in a row, such that each number, other than the left-most, is within one of some number to the left of it.
Labels: mathschallenge
posted by Zaux at
6:00 PM
2 Comments:
512
Cam
Sequencing the digits
Fix the first number of the digit.
Count the # for each digit.
after selecting a first number the elements following must be either bigger or smaller in a series
e.g
start with 4.
bigger elements are chosen from
B=5,6,7,8,9 in sequence
smaller elements are chosen from
S=3,2,1,0 in sequence
the number of possible permutations of a group with 2 different elements repeating A, B times is:
P=N!/(A!B!)
e.g. for 4 we represent the 9 digits following 4 as the permutations of
BBBBBSSSS
P=9!/(5!4!)=126
For 0 we have BBBBBBBBB P=9!/(9!*0!)=1
For 1 we have SBBBBBBBB P=9!/(8!*1!)=9
For 2 we have SSBBBBBBB P=9!/(7!*2!)=36
For 3 we have SSSBBBBBB P=9!/(6!*3!)=84
For 4 we have SSSSBBBBB P=9!/(5!*4!)=126
For 5 we have SSSSSBBBB P=9!/(4!*5!)=126
For 6 we have SSSSSSBBB P=9!/(3!*6!)=84
For 7 we have SSSSSSSBB P=9!/(2!*7!)=36
For 8 we have SSSSSSSSB P=9!/(1!*8!)=9
For 9 we have SSSSSSSSS P=9!/(0!*9!)=9
1+9+36+84+126+126+84+36+9+1=512
Answer: 512
Cam
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