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Combinatorics Through Guided Discovery, 2004a

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84 4. Generating Functions<br />

the generating function for partitions whose Young diagram fits into an m<br />

by n rectangle, as in Problem 167. This generating function has significant<br />

analogs to the binomial coefficient ( m+n<br />

n<br />

), and so it is denoted by [m+n n<br />

] q .It<br />

is called a q-binomial coefficient.<br />

(a) Compute [ 4 2 ] q = [2+2 2 ] q . (h)<br />

(b) Find explicit formulas for [ n 1 ] q and [ n<br />

n−1 ] q . (h)<br />

(c) How are [ m+n<br />

n<br />

] q<br />

asking how [ r s ] q and [ r<br />

r−s ] q<br />

and [m+n<br />

n<br />

] related? Prove it. (Note this is the same as<br />

q<br />

are related.) (h)<br />

(d) So far the analogy to ( m+n )<br />

n<br />

is rather thin! If we had a recurrence<br />

like the Pascal recurrence, that would demonstrate a real analogy. Is<br />

[ m+n<br />

n<br />

] q = [m+n−1 n−1<br />

] q + [m+n−1 ]<br />

n q<br />

?<br />

(e) Recall the two operations we studied in Problem 171.<br />

(i) The largest part of a partition counted by [ m+n<br />

n<br />

] is either m or<br />

q<br />

is less than or equal to m − 1. In the second case, the partition<br />

fits into a rectangle that is at most m − 1 units wide and at most<br />

n units deep. What is the generating function for partitions of<br />

this type? In the first case, what kind of rectangle does the<br />

partition we get by removing the largest part sit in? What is the<br />

generating function for partitions that sit in this kind of rectangle?<br />

What is the generating function for partitions that sit in this kind<br />

of rectangle after we remove a largest part of size m? What<br />

recurrence relation does this give you?<br />

(ii) What recurrence do you get from the other operation we studied<br />

in Problem 171?<br />

(iii) It is quite likely that the two recurrences you got are different.<br />

One would expect that they might give different values for [ m+n<br />

n ] q .<br />

Can you resolve this potential conflict? (h)<br />

(f) Define [n] q to be 1+q + ···+ q n−1 for n > 0 and [0] q =1. We read<br />

this simply as n-sub-q. Define [n]! q to be [n] q [n − 1] q ···[3] q [2] q [1] q .We<br />

read this as n cue-torial, and refer to it as a q-ary factorial. Show that<br />

[ ] m + n<br />

n<br />

q<br />

= [m + n]! q<br />

[m]! q [n]! q<br />

.<br />

(h)<br />

(g) Now think of q as a variable that we will let approach 1.<br />

explicit formula for<br />

Find an<br />

(i) <br />

q→1<br />

[n] q .

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