(a) Count the number of quadruples (a, b, c, d) of nonnegative integers such that 0<= a < b < c < d <= 12.

(b) For this part, we want to count the number of quadruples (a, b, c, d) of nonnegative integers such that 0 <= a <= b <= c <= d <= 12.

Here, some of a, b, c and d can be equal to each other, so the answer will be different from part (a). Each value a, b, c, d must be between 0 and 12 inclusive. One idea is to count how many times each number appears.

(c) In general, find the number of k-tuples (a_1, a_2, ... a_k) of nonnegative integers such that

0 <= a_1 <= a_2 <= a_3 <= . . . <= a_k <= n.

DeSTrOYer303 Apr 7, 2020

#1

#4**0 **

I checked the answer to that question and the answerer did not consider 0 as a valid digit, therefore all of his/her answers were based off of 12 instead of 13.

DeSTrOYer303
Apr 7, 2020

#2**+1 **

(a): This seems complicated, yet I think 13 C 4 = 715 ways if a is the least, b is the second to least, etc.

Yes, you're right...

tertre Apr 7, 2020

#3**+1 **

There would be 13 integers to choose from right?

So it would be 13 C 4= 715

HELPMEEEEEEEEEEEEE
Apr 7, 2020

#7**0 **

I did get 715 for the first part, and I did not have trouble with that. For part b, I just don't know whether to approach this problem with casework or some sort of counting technique like stars and bars. One of the hints for part b was to approach the problem by designating variables.

DeSTrOYer303
Apr 7, 2020

#8**0 **

For b., if a were 13 then there is one arrangement where all of them are 13, if a were 12 then there are 4 arragements... I wonder if there is a pattern.

HELPMEEEEEEEEEEEEE
Apr 7, 2020

#11**0 **

I'm not seeeing any pattern here, so we should probably look for something else...

HELPMEEEEEEEEEEEEE
Apr 7, 2020

#13**0 **

That is a great observation! If that is the pattern, then you should be able to solve the problem from there!

HELPMEEEEEEEEEEEEE Apr 7, 2020