- UID
- 1082718
- 在线时间
- 小时
- 注册时间
- 2015-1-22
- 最后登录
- 1970-1-1
- 主题
- 帖子
- 性别
- 保密
|
134 题
If 4 people have exactly 1 sibling, there are 2 pairs of siblings in the 4 people
Say A B C and D, each have exactly 1 sibling. That implies, A has 1 sibling (say B), so B's sibling is A
Similarly, C and D are siblings
Now if 3 people (say E F and G) each have 2 siblings, then E must have F and G both as siblings
F must have E and G both as siblings, and same for G too.
So we have 2 pairs of 2 siblings and 1 triplet of 3 siblings
A-B
C-D
E-F-G
ways of selecting 2 out of 7 is = 21 ways
ways of NOT selecting a sibling pair = Total ways - ways of always getting a pair
ways of getting a sibling pair :
A-B = 1way
C-D = 1 way
2 out of 3 (E - F - G) = 3 ways
so 5 ways we always get a sibling pair, remaining 21-5 ways we dont get a sibling pair
ans is 16/21 |
|