Question

A committee of \(8\) members is to be formed from \(5\) teaching staff, \(4\) office staff and \(6\) students so as to include at least two from each category. Then the total number of ways of forming the committee is

• A. \(4100\)
• B. \(3950\)
• C. \(3200\)
• D. \(3500\) \bigskip

Hint:

For committee formation problems with minimum restrictions, first distribute the required number of members among the categories, then count each case using combinations and add the results.

Answer 1

The Correct Option is A

Concept: The committee must contain \(8\) members selected from three categories: \[ \text{Teaching Staff }(T)=5,\quad \text{Office Staff }(O)=4,\quad \text{Students }(S)=6. \] Since at least \(2\) members must be chosen from each category, we distribute \(8\) members among the three categories and then use combinations.

Step 1: Find all possible distributions of \(8\) members with at least \(2\) from each category. Let the numbers selected from the three categories be \[ (t,o,s). \] Then \[ t+o+s=8, \] with \[ t\ge 2,\quad o\ge 2,\quad s\ge 2. \] Possible distributions are: \[ (2,2,4), \] \[ (2,3,3), \] \[ (2,4,2), \] \[ (3,2,3), \] \[ (3,3,2), \] \[ (4,2,2). \]

Step 2: Count committees for each distribution. For \((2,2,4)\), \[ \binom{5}{2}\binom{4}{2}\binom{6}{4} = 10\times 6\times 15 = 900. \] For \((2,3,3)\), \[ \binom{5}{2}\binom{4}{3}\binom{6}{3} = 10\times 4\times 20 = 800. \] For \((2,4,2)\), \[ \binom{5}{2}\binom{4}{4}\binom{6}{2} = 10\times 1\times 15 = 150. \] For \((3,2,3)\), \[ \binom{5}{3}\binom{4}{2}\binom{6}{3} = 10\times 6\times 20 = 1200. \] For \((3,3,2)\), \[ \binom{5}{3}\binom{4}{3}\binom{6}{2} = 10\times 4\times 15 = 600. \] For \((4,2,2)\), \[ \binom{5}{4}\binom{4}{2}\binom{6}{2} = 5\times 6\times 15 = 450. \]

Step 3: Add all the cases. \[ 900+800+150+1200+600+450 = 4100. \]

Step 4: Write the final answer. \[ \boxed{4100} \]