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  Choose the correct choice:
A) 38 B) 46 C) 72 D) 86 E) 102
Without limitations, 5 children can be seated in 5!=120 ways. Find the number of ways to seat the 5 children so that the siblings DO sit together. The siblings can be regarded as one unit so there are 4! combinations. But within this unit the siblings can sit in two different ways. So the number of ways to seat the 5 children so that the siblings DO sit together is 4!∗2=48. Thus, the number of combinations in which the siblings DO NOT sit together is  120 - 48= 
<gadget id="calculator"> 120 - 48</gadget>
<output>72</output>
72.   Answer: 72
<result>C</result>
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