Question

There are 3 students from section A, 5 from section B and 4 from section C. In how many ways can they occupy 6 seats such that the leftmost seat is occupied by a student of section A and the remaining seats are occupied by 3 students of section B and 2 students of section C?

• A. 10800
• B. 14400
• C. 18000
• D. 21600

Hint:

Fix the special condition (which section sits leftmost) first, then handle the remaining seats. When a group of seats will be shared by two different sections, decide how many seats each section gets before deciding who specifically sits in them.

Answer 1

The Correct Option is D

Step 1: Choose the student for the leftmost seat from section A.
There are 3 students in section A, and the leftmost seat must be occupied by a student from section A.
Number of ways = \(3\)
Step 2: Select students from section B and section C.
We need to choose 3 students from section B out of 5:
\[ \binom{5}{3} = 10 \]
We need to choose 2 students from section C out of 4:
\[ \binom{4}{2} = 6 \]
Step 3: Arrange the selected students in the remaining 5 seats.
The selected 5 students (3 from B and 2 from C) can be arranged in the remaining 5 seats in:
\[ 5! = 120 \]
Step 4: Multiply all possible cases.
\[ 3 \times 10 \times 6 \times 120 = 21600 \]
Final Answer:
\[ \boxed{21600} \]

Answer 2

Concept:

• Instead of selecting the whole group of 5 students first and then arranging all of them together, decide which of the 5 non-leftmost seats go to Section B and which go to Section C, then separately select and arrange students into each set of designated seats.
• Selecting $r$ distinct people out of $n$ and arranging them, in order, into $r$ specific seats can be counted directly with the permutation formula $^nP_r = \dfrac{n!}{(n-r)!}$, without splitting it into a separate "choose" step and "arrange" step.

Step 1: Fill the leftmost seat with a Section A student.
Any of the 3 Section A students can sit there: 3 ways.

Step 2: Decide which of the remaining 5 seats belong to Section B.
Choose 3 of the 5 remaining seats to be "B seats" (the other 2 automatically become "C seats"): $\binom{5}{3} = 10$ ways.

Step 3: Fill the 3 B-seats, in order, with 3 of the 5 Section B students.
$^5P_3 = 5 \times 4 \times 3 = 60$ ways.

Step 4: Fill the remaining 2 C-seats, in order, with 2 of the 4 Section C students.
$^4P_2 = 4 \times 3 = 12$ ways.

Step 5: Multiply every independent choice together.
$3 \times 10 \times 60 \times 12 = 21600$

Final Answer: $21600$