CSE390D Notes for Friday, 11/8/24
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We are going to fill in a table with four possibilities for picking m
items from a set of n items...seen 2 possibilities so far:
w/o repetition with repetition
ordered P(n, m) ??
unordered C(n, m) ??
Review of those:
38 students in a class
how many ways to seat them in 38 chairs?
38!
how many ways to fill front row of 5 chairs?
P(38, 5) = 38!/33!
how many ways to pick 6 winners of a candy bar?
C(38, 6) = 38!/(6! * 32!)
how many ways to pick 32 losers of a candy bar?
C(38, 32) = 38!/(32! * 6!)
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this example includes idea of choosing positions and is good lead-in to
binomial coefficients:
How many paths are there from (0, 0) to (x, y) (x/y in N), if paths involve
either moving up 1 or right 1?
e.g., get to (3, 2):
some paths: RRRUU, UURRR, RURUR
answer: choose positions of either U's or R's, so C(x + y, x) or C(x + y, y)
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binomial coefficients
(x + y) ^ n
coefficient of x^m y^(n-m) is C(n, m) = C(n, n - m) = (n choose m)
a counting intuition:
(x + y)^3 = (x + y)(x + y)(x + y)
= (xx + xy + yx + yy)(x + y)
= (xxx + xxy + xyx + xyy + yxx + xyx + yyx + yyy)
how many x^2y? three: xxy, xyx, yxx
why?
given 2x's & 1 y, there are 3 possible sequences
in general, given a sequence of length n with m x's, how many sequences?
(n choose m) -- pick the m positions of the x's
notice that sum(k = 0 to n) C(n, k) = 2^n because 2^n = (1 + 1)^n
Pascal's triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Pascal's triangle rule:
(n choose m) = (n-1 choose m-1) + (n-1 choose m)
why?...think about Joe (choose m winners, Joe is either a winner or not)
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Back to the table:
w/o repetition with repetition
ordered P(n, m) ??
unordered C(n, m) ??
ordered with repetition: r^n
list "employee of the month" for a company of 35 employees
how many outcomes for 12 months? 35^12
unordered with repetition:
stars & bars
8 contest winners can pick any of three prizes (a, b, c)
how many prize orders are possible (# of a, b, c)?
a...a|b...b|c...c
stars & bars, (8+2 choose 2)
In general (picking m items from a set of n items)
w/o repetition with repetition
ordered P(n, m) n^m
unordered C(n, m) P(m+n-1, n-1)
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Stuart Reges
Last modified: Fri Nov 8 12:56:15 PST 2024