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11 Jan 2022, 08:00
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:57) correct
54%
(02:38)
wrong
based on 76
sessions
History
Date
Time
Result
Not Attempted Yet
From 8 people, a team of 4 needs to be selected. How many ways are possible if two particular people want to be together, while two other people do not want to be together?
(A) 9
(B) 14
(C) 15
(D) 17
(E) 23
Are You Up For the Challenge: 700 Level Questions
(A) 9
(B) 14
(C) 15
(D) 17
(E) 23
Are You Up For the Challenge: 700 Level Questions
How many ways to create 4 person team from 8 absent restrictions:
8!/4!4!= 70
Call the two together AB and two apart EF
Subtract from 70 combinations of AnotB and BnotA:
A fills 1 slot. 6 remaining, excluding B, select 3
6!/3!3!= 20
Same answer for BnotA = 20
Subtract combinations with EF together. Need to fill 2 from 6:
6!/2!4!= 15
ADD back double counts
AEF plus 1 from other 4= 4
BEF plus 1 from other 4=4
Total:
70-20-20-15+4+4 = 23
Another way, not involving subtractions:
ABC:need 1 from EFGH = 4 ways
ABD:need 1 from EFGH = 4 ways
AB: need 2 from EFGH = 6 ways
C: need 3 from EFGH = 4 ways
D: need 3 from EFGH = 4 ways
EFGH:need 4 from EFGH = 1 way
Total ways: 23
Posted from my mobile device
8!/4!4!= 70
Call the two together AB and two apart EF
Subtract from 70 combinations of AnotB and BnotA:
A fills 1 slot. 6 remaining, excluding B, select 3
6!/3!3!= 20
Same answer for BnotA = 20
Subtract combinations with EF together. Need to fill 2 from 6:
6!/2!4!= 15
ADD back double counts
AEF plus 1 from other 4= 4
BEF plus 1 from other 4=4
Total:
70-20-20-15+4+4 = 23
Another way, not involving subtractions:
ABC:need 1 from EFGH = 4 ways
ABD:need 1 from EFGH = 4 ways
AB: need 2 from EFGH = 6 ways
C: need 3 from EFGH = 4 ways
D: need 3 from EFGH = 4 ways
EFGH:need 4 from EFGH = 1 way
Total ways: 23
Posted from my mobile device
20 Jan 2022, 10:42
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Let's say A and B must stay together and conversely C and D don't want to end up being in the same group
So let's start with groups containing A e B
We have:
A, B, 6C2 = 15
Now we have to subtract the
A, B, C, D selection
So we end up with a (sub)total of 14 (15-1)
Now we opt for neither A or B being part of the selected group of four
6C4 = 15
As we did before we do have to account for groups containing C and D together, which need to be subtracted from the result we have just found
C, D, 4C2 = 6
(note that we are choosing 2 out of 4 instead of 6 since we are in the sub-case where A and B cannot be part of the selection)
So we end up with a (sub)total of 9 (15-6)
Grand total = 23
So let's start with groups containing A e B
We have:
A, B, 6C2 = 15
Now we have to subtract the
A, B, C, D selection
So we end up with a (sub)total of 14 (15-1)
Now we opt for neither A or B being part of the selected group of four
6C4 = 15
As we did before we do have to account for groups containing C and D together, which need to be subtracted from the result we have just found
C, D, 4C2 = 6
(note that we are choosing 2 out of 4 instead of 6 since we are in the sub-case where A and B cannot be part of the selection)
So we end up with a (sub)total of 9 (15-6)
Grand total = 23
