數學題(Combination)#9 10分

38 The r + 1 th term is C(n,r), The r th term is C(n,r - 1)

The r + 10 th term is C(n,r + 9), The r + 11 th term is C(n,r + 10)


n!/[r!(n - r)!] = 2n!/[(r - 1)!(n - r + 1)!]...(1)


n!/[(r + 9)!(n - r - 9)!] = 2n!/[(r + 10)!(n - r - 10)!]...(2)

So,

2r = n - r + 1...(3)

2(n - r - 9) = r + 10...(4)

Solve it,

4r - 20 = r + 10

3r = 30

r = 10 => n = 29

11(a) A->D->E->G ; A->D->F->G ; A->B->C->F->G

3 routes

(b) 3 * 3 = 9 routes

(c) E->D ; F->D ; G->E->D ; G->F->D

4 return routes

38.
general (x+y)^n = summation of nCw [x^(n-w)] [y^w]

(1+x)^n
= summation of nCw [1^(n-w)] [x^w]
(r+1)th term => w = r
the coefficient at (r+1)th term = nCr
rth term => w = r-1
the coefficient at rth term = nCr-1
(r+10)th term => w = r+9
the coefficient at rth term = nCr+9
(r+11)th term => w = r+10
the coefficient at rth term = nCr+10
nCr = 2 nCr-1
n! / [r!(n-r)!] = 2(n!) / [(r-1)!(n-r+1)!]
1/ r = 2/ (n-r+1)
2r = n-r+1 => 3r = n + 1 ... (1)
nCr+9 = 2 nCr+10
n! / [(r+9)!(n-r-9)!] = 2(n!) / [(r+10)!(n-r-10)!]
1/ (n-r-9) = 2/ (r+10)
r + 10 = 2n-2r-18 => 3r = 2n - 28 ...(2)
sub (1) into (2)
n + 1 = 2n - 28
n = 29
=> r = 10

11a.
(without backtracking supposes no city to be reached twice)
A>B>C>F>D>E>G
A>B>C>F>G
A>B>D>E>G
A>B>D>F>G
so it should be 4 routes from city A to city G
11b.
from city A to city G has 4 possible routes
from city G to city A has 4 possible routes
so total is 4x4 = 16 possible routes from city A to city G and back to city A
11c.
consider return routes only
there are 3 possible routes on return travel through city D