How many four digit numbers are there where the digits are ascending from left to right?
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The first of the 8 digits can’t be a 0, for then the number wouldn’t be an 8-digit number. So the 8 digits are 8 of {1, 2, 3, 4, 5, 6, 7, 8, 9}, ascending from left to right. There’s a choice of 9 digits to be the one that is absent, so there are 9 such 8-digit numbers.
Since we can’t begin with a zero, we must CHOOSE any 4 of the remaining 9 digits. As I tell my students, a clear problem statement contains the solution: “9 choose 4.” C(9,4) = 126.
Nice review problem; I’m swiping it (and it’s 3- and 5-digit cousins) for my lectures and quizzes. Thanks!
Here’s a variation for the more sporting puzzle-solver. I ask my students to find the probability of being dealt a flush in a hand of 5-card stud. There’s no pesky zero, but there are 4 suits.
Well, there are C(13,5) = 1,287 ways to get all spades, or 4 times that = 5,148 ways to get a flush in any suit. Since there are C(52,5) ways to get a 5 card hand, the probability of getting a flush is 13*12*11*10*9/52*51*50*49*48.
If we’re playing Think Again old school, let me pose this. In a game with 6 players, is it easy to compute the chances that at least one player will have a flush?
February 23rd, 2010 at 2:37 am
The first of the 8 digits can’t be a 0, for then the number wouldn’t be an 8-digit number. So the 8 digits are 8 of {1, 2, 3, 4, 5, 6, 7, 8, 9}, ascending from left to right. There’s a choice of 9 digits to be the one that is absent, so there are 9 such 8-digit numbers.
February 23rd, 2010 at 6:11 am
Thank you for the solution!
Ooops! It should have been ‘four digit numbers’ not ‘eight digit numbers’. The problem has been corrected.
February 23rd, 2010 at 7:16 pm
Since we can’t begin with a zero, we must CHOOSE any 4 of the remaining 9 digits. As I tell my students, a clear problem statement contains the solution: “9 choose 4.” C(9,4) = 126.
Nice review problem; I’m swiping it (and it’s 3- and 5-digit cousins) for my lectures and quizzes. Thanks!
February 23rd, 2010 at 7:22 pm
Here’s a variation for the more sporting puzzle-solver. I ask my students to find the probability of being dealt a flush in a hand of 5-card stud. There’s no pesky zero, but there are 4 suits.
February 24th, 2010 at 11:46 am
Well, there are C(13,5) = 1,287 ways to get all spades, or 4 times that = 5,148 ways to get a flush in any suit. Since there are C(52,5) ways to get a 5 card hand, the probability of getting a flush is 13*12*11*10*9/52*51*50*49*48.
If we’re playing Think Again old school, let me pose this. In a game with 6 players, is it easy to compute the chances that at least one player will have a flush?