# Combination Question

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#1
Question:
A girl wants to ask eight friends to tea, but there is only room for four of them. In how many ways can she choose whom to invite if two of them are sisters and must not be separated? (Consider two cases a) when both sisters are invited b) when neither sister is invited)

a) when both sisters are selected;
S1 S2 F1 F2 F3 F4 F5 F6

S1 S2 _ _

The two spaces can be selected; 6C2 = 6!/(6-2)!(2)! 15 ways

b) when neither sister is invited;

S1 S2 F1 F2 F3 F4 F5 F6

_ _ _ _
The four spaces are available for Friends not sisters
thus there are 6 friends to choose from for the 4 spaces;
6C4 = 6!/(6-4)!(4!) = 15 ways

thank you for helping
0
1 year ago
#2
Those two sets sound like they're mutually exclusive and complete so it should be simple to combine?
(Original post by bigmansouf)
Question:
A girl wants to ask eight friends to tea, but there is only room for four of them. In how many ways can she choose whom to invite if two of them are sisters and must not be separated? (Consider two cases a) when both sisters are invited b) when neither sister is invited)

a) when both sisters are selected;
S1 S2 F1 F2 F3 F4 F5 F6

S1 S2 _ _

The two spaces can be selected; 6C2 = 6!/(6-2)!(2)! 15 ways

b) when neither sister is invited;

S1 S2 F1 F2 F3 F4 F5 F6

_ _ _ _
The four spaces are available for Friends not sisters
thus there are 6 friends to choose from for the 4 spaces;
6C4 = 6!/(6-4)!(4!) = 15 ways

thank you for helping
0
1 year ago
#3
For a) I got 12
For b) don't know what is going on
Last edited by As.1997; 1 year ago
0
1 year ago
#4
I learnt these without the formula. It helps you to actually understand whatâ€™s going on and why.
0
1 year ago
#5
15 = 6C2 = 6C4
Sounds right for both parts.
You're interested in combinations, i.e. no repitition and order unimportant.
(Original post by As.1997)
For a) I got 12
For b) don't know what is going on
0
1 year ago
#6
(Original post by mqb2766)
15 = 6C2 = 6C4
Sounds right for both parts.
You're interested in combinations, i.e. no repitition and order unimportant.
Itâ€™s very strange to me how if youâ€™re limited and can only vary the 2 invites with 6 people that is still equal to having the 4 invites with 6 people. Intuitively I would have thought you would be able to have many more combinations with 6C4.
0
1 year ago
#7
sure, but think about the number of combinations left out. Its a mirror of the other question part, so they must be the same. In general, nCr = nC(n-r), have a bit of a read about binomial coefficients and Pascal's triangle to find out more.
(Original post by Maximus 190)
Itâ€™s very strange to me how if youâ€™re limited and can only vary the 2 invites with 6 people that is still equal to having the 4 invites with 6 people. Intuitively I would have thought you would be able to have many more combinations with 6C4.
0
#8
(Original post by mqb2766)
Those two sets sound like they're mutually exclusive and complete so it should be simple to combine?
thank you
0
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