Bunuel wrote:
If among 5 children there are 2 siblings, in how many ways can the children be seated in a row so that the siblings do not sit together?
A. 38
B. 46
C. 72
D. 86
E. 102
1. "seated in a row" = order matters, permutation
2. replacement? No
3. different objects? Yes
4. different groups? No (row = 1 group)
Slot method: A1 first slot = 3
A1 _ A2 _ _
A1 _ _ A2 _
A1 _ _ _ A2
A1 2nd slot = 2
_ A1 _ A2 _
_ A1 _ _ A2
A1 3rd slot = 1
_ _ A1 _ A2
(3+2+1) or 3P1 = 6 * 2 (doubled for A2) = 12 ways to arrange the 2 siblings
3P1 = 3*2*1 ways to arrange the other children
12*6 = 72
Another way to think about it ... ways*choices: Step 1) Choices of objects per slot, 2P1 for siblings * 3P1 for others = 12
_2_ * _3_ * _1_* _2_* _1_ = 12
Step 2) Ways to position the siblings among 5 slots so they don't sit together, 3C1 * 2 = 6
(leftmost slot, 2nd to leftmost = 2 and middle are viable --> 3 * 2 for each sibling = 6 )
ways * choices = 6 ways * 12 obj choices * 1 grp choice = 72