[Guest]

A committee of five is to be chosen from a group of nine people.

Determine the total number of ways this can be accomplished, given that both these conditions must be satisfied.

(i) Ted and Chuck must serve together or not at all, and:

(ii) Hannah and Cindy refuse to serve with each other.

Edited January 13, 2010 by K Sengupta

[roolstar]

A committee of five is to be chosen from a group of nine people.

Determine the total number of ways this can be accomplished, given that:

(i) Ted and Chuck must serve together or not at all, and:

(ii) Hannah and Cindy refuse to serve with each other.

I have no idea how to write the combinations notation in this box so explenation first:

7C5 means the number of combinations of 5 people in a group of 7

(i) Both in + both out = 7C3 + 7C5 = 35 + 21 = 56 ways

(ii) 2xonly one in + none of them in = 2 x 7C4 + 7C5 = 2x35 + 21 = 91 ways

(i) and (ii)

= None of the four characters in + the two men in none of the girls in + two men in one girl in x 2 + One girl in the two guys out x2

= 5C5 + 5C3 + 5C2 x 2 + 5C4 x 2 = 1 + 10 + 10 x 2 + 5 = 36 ways

I hope I didn't make any mistakes rushing into this...

Edited January 13, 2010 by roolstar

[Guest]

I think it can be done in 41 ways

When either hannah or Cindy is in:

2 x (5C2 + 5C4) = 30

5C2 when both Ted and Chuck are in and 5C4 when neither Ted nor Chuck is in

When neither hannah nor Cindy is in

5C3 + 1 = 11

5C3 when Ted and chuck are in and 1 when neither Ted nor Chuck is in

Total number of ways = 30+11 = 41

[Guest]

I think it can be done in 41 ways

When either hannah or Cindy is in:

2 x (5C2 + 5C4) = 30

5C2 when both Ted and Chuck are in and 5C4 when neither Ted nor Chuck is in

When neither hannah nor Cindy is in

5C3 + 1 = 11

5C3 when Ted and chuck are in and 1 when neither Ted nor Chuck is in

Total number of ways = 30+11 = 41

41

[Guest]

I think it can be done in 41 ways

When either hannah or Cindy is in:

2 x (5C2 + 5C4) = 30

5C2 when both Ted and Chuck are in and 5C4 when neither Ted nor Chuck is in

When neither hannah nor Cindy is in

5C3 + 1 = 11

5C3 when Ted and chuck are in and 1 when neither Ted nor Chuck is in

Total number of ways = 30+11 = 41

I split the choices out as follows:

When choosing Hannah or Cindy:

2 (Hannah or Cindy) * 1 (Ted or Chuck) * 1 (Chuck or Ted) * 5C2 = 2*10 = 20

2 (Hannah or Cindy) * 5C4 = 2*5 = 10

(and now I see where you got the (5C2 + 5C4) part...)

When not choosing Hannah or Cindy:

1 (Ted or Chuck) * 1 (Chuck or Ted) * 5C3 = 10

When choosing no named parties:

5C5 = 1

Total = 20 + 10 + 10 + 1 = 41 combinations

[Guest]

Interesting twist, to get you thinking...

The committee has 5 DIFFERENT positions: President, Vice President, Secretary, Treasurer, and Member at Large. Therefore the committee is technically different if Ted is President and Chuck is Vice President, vs. Chuck being President and Ted Vice President (if all other members are still the same). The same rules apply:

(i) Ted and Chuck must serve together or not at all, though it doesn't matter what position each holds, and

(ii) Hannah and Cindy refuse to serve with each other.

Still selecting from the same 9 people, now how many different ways can this be accomplished?

Edited January 13, 2010 by LMcNair27

[Guest]

I am thinking 41.

if you use Ted and Chuck and one of the girls there are 10 possible combinations of other members. do this twice for each of the girls. so that's 20. Then do just Ted and Chuch with neither girl which gives you another 10. then there is only one girl with no Ted or Chuck and another 4 members which gives you 5 possible combinations and multiply that by 2 for either girl for another 10, then there is the one combination if none of the named people are selected. and the total is 41.

[Guest]

Seems kind of pointless now but

41

Keep them coming Sengupta!

[Guest]

When Ted and Chuck are both in, with either Cindy or Hannah are in, only 2

of the other 5 can be used, giving 20 combinations, 10 for Cindy in and 10 for

Hannah in.

When Ted and Chuck are both out, and either Cindy or Hannah are in, 4 out

of the remaining 5 can be used, resulting in 10 combinations, 5 for Cindy in

and 5 for Hannah in.

There are no combinations for the remaining 5 by themselves, since either

Cindy or Hannah has to serve.

The total combinations is 30.

[Guest]

There are no combinations for the remaining 5 by themselves, since either

Cindy or Hannah has to serve.

This statement is false. The constraint was:

(ii) Hannah and Cindy refuse to serve with each other.

If neither one is serving, then they are not serving with each other.

[Guest]

Interesting twist, to get you thinking...

The committee has 5 DIFFERENT positions: President, Vice President, Secretary, Treasurer, and Member at Large. Therefore the committee is technically different if Ted is President and Chuck is Vice President, vs. Chuck being President and Ted Vice President (if all other members are still the same). The same rules apply:

(i) Ted and Chuck must serve together or not at all, though it doesn't matter what position each holds, and

(ii) Hannah and Cindy refuse to serve with each other.

Still selecting from the same 9 people, now how many different ways can this be accomplished?

NEWBIE ALERT!!!!!!!! COULD BE LONG BY A CoUPLA THOUSAND LIGHT YEARS!!!!!!!!!!!

I'm guessing

1008000