In the explanation (1st method) - for the case starting with 3, you deduced that there are 9c2 ways for 2,1. The question is: I understand 9c2 is the number of ways to select 2 places/positions out of 9 total positions. right? without taking into account the order (2 must come before 1). So does that still apply? I know 9c2 i right but I don't know if this works for both: a) selecting any 2 positions for 1 and 2 (without order) and b) selecting 2 positions for 2 numbers but one of the numbers must be before the other. May you explain this point please
If you choose two positions, one of them must come before the other. Finding the number of combinations just implies that the 2 will occupy the first position relative to the two that are chosen.
Could we switch minds on February 19th?
The problem states that the list of 10 number contains the first 10 integers, so no number can be included twice.
Loved the explanation with multiple approaches. Thanks.
In the second solution, how does 2^9 include the lists that start with other numbers other than 5?
I am taking the AMC 10 tomorrow, lets switch brains for a day.
How was it??
I like the way you explain. thanks
In the explanation (1st method) - for the case starting with 3, you deduced that there are 9c2 ways for 2,1. The question is: I understand 9c2 is the number of ways to select 2 places/positions out of 9 total positions. right? without taking into account the order (2 must come before 1). So does that still apply? I know 9c2 i right but I don't know if this works for both: a) selecting any 2 positions for 1 and 2 (without order) and b) selecting 2 positions for 2 numbers but one of the numbers must be before the other. May you explain this point please
If you choose two positions, one of them must come before the other. Finding the number of combinations just implies that the 2 will occupy the first position relative to the two that are chosen.
Billy Black thank you 👍
How do we know that we can`t repeat the number?
wait wouldnt the 2^9 inlcude going down 9 times?
10,9,8,...,2,1 fulfils the conditions
@@balladenumberone oh thanks
THANKS SO MUCH!
excellent
why does 9c1 + 9c2 + 9c3 + ... + 9c9 = (1+1)^2??
Try expanding (1+1)^9 using the binomial theorem.
wow thanks! is there a proof for this? Is this used often? I'm taking the AMC 10 tomorrow and I don't know if I'll make AIME
Binomial theorem
There is an identity where nC1 + nC2 + … +nCn = 2^n
In the sequence which has 1 for the first term,
why can`t 1 appear again in the 3rd term?