A group of 6 students - Alice, Bob, Charlie, David, Eve, and Frank - are to be seated in a row. Alice and Bob must sit together, and David cannot sit at either end of the row. How many different seating arrangements are possible?
Options
A
144
B
192
C
240
D
288
Answer & Analysis
Answer
A
Analysis
Question Analysis
This question involves multiple permutation constraints.
The main focus is on using the “grouping” method for adjacent elements and the “restricted position” approach for elements with specific placement limitations.
It requires applying basic permutation formulas while considering these special conditions.
Key Concept Explanation
Grouping adjacent elements: When two elements must be together, we consider them as a single entity for the initial permutation calculation.
The number of ways to arrange the elements within the group is also factored in.
Restricted position: For an element with a restricted position (like David not sitting at the ends), we account for the available positions and adjust the permutation calculations accordingly.
Step - by - Step Solution
Treat Alice and Bob as one unit:
The number of ways to arrange Alice and Bob within this unit is (either Alice - Bob or Bob - Alice).
Now, considering this unit along with Charlie, Eve, Frank, and the “David” element, we have a total of 5 units to arrange.
But David cannot be at either end.
So, for the first position, we have 4 choices (the Alice - Bob unit, Charlie, Eve, or Frank).
After placing the first unit, for the last position, we have 3 remaining non - David choices.
And for the middle 3 positions, the remaining 3 units can be arranged in ways.
The number of ways to arrange these 5 units with the given David's restriction is .
Calculate
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