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# hi what do i do here

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How many subsets of the set {1,2,...,11} have median 6?

Feb 28, 2022

#1
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My only way is to try to list them; this is nasty!

Case 1:  the subset contains the number 6

A)  You choose five numbers lower than 6 (1 way)

and five numbers larger than 6 (1 way) for a total of 1 way

B)  You choose four numbers lower than 6 (5 ways)

and four numbers larger than 6 (5 ways) for a total of 5 x 5 = 25 ways

C)  You choose three numbers lower than 6 (10 ways)

and three numbers larger than 6 (10 ways) for a total of 10 x 10 = 100 ways

D)  You choose two numbers lower than 6 (10 ways)

and to numbers larger than 6 (10 ways) for a total of 10 x 10 = 100 ways

E)  You choose one number lower than 6 (5 ways)

and one number larger than 6 (5 ways) for a total of 5 x 5  =  26 ways

F)  You choose no number lower than 6 (1 way)

and no number larger than 6 (1 way) for a total of 1 x 1 = 1 way

Now, for the really nasty part, you don't choose the number 6 -- when you do this,

the largest number below 6 and the smallest number above 6 must have a mean

of 6.

A)  You choose five numbers lower than 6 (1 way)

and five numbers larger than 6 (1 way) for a total of 1 way

B)  You choose four numbers lower than 6 and four numbers larger than 6 (10 ways)

C)  You choose three numbers lower than 6 and three numbers larger than 6 (46 ways)

D)  You choose two numbers lower than 6 and two numbers larger than 6 (30 ways)

E)  You choose one number lower than 6 and one number larger than 6 (5 ways)

Hopefully, I counted correctly --  if not, maybe someone can correct me ...

Feb 28, 2022
#3
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Hi Geno, let's see if i get the same :)

How many subsets of the set {1,2,...,11} have median 6?

1 2 3 4 5    6    7 8 9 10 11

The 6 is included

1+(5C4)^2+(5C3)^2+(5C2)^2+(5C1)^1+1

= 1+5^2    +10^2      +10^2    + 5^2     +1

= 1+25+100+100+25+1

=252

6 not included but 5 and 7 both included

(4C4)^2+(4C3)^2+(4C2)^2+(4C1)^2+4C0)^2

=1         +4^2          +6^2      +4^2     +1

=1+16+36+16+1

=70

5,6, and 7 not included but 4 and 8 inlcuded

(3C3)^2+(3C2)^3+(3C1)^2+(3C0)^2

=1+9+9+1

=20

4,5,6,7,8 not included but 3 and 9 included

(2C2)^2+(2C1)^2+(2C0)^2

=1+4+1

=6

3,4,5,6,7,8,9 not included but  2 and 10 included

1+1

= 2

2,3,4,5,6,7,8,9,10 not included but 1 and 11 included

=1

252+70+20+6+2+1 = 351

Feb 28, 2022