From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?
The number of ways to choose a chairman and a vice chairman from 8 persons is 56.
Since we need to choose a Chairman and a Vice-Chairman from 8 persons, and one person cannot hold more than one position, this is a permutation problem where order matters. The two positions are distinct, so arranging persons in order is important.
Number of ways = ⁸P₂ = 8!/(8-2)! = 8!/6! = = 56 ways
Alternatively, we can solve this using the multiplication principle: The Chairman can be chosen in 8 ways (from 8 persons). After choosing the Chairman, the Vice-Chairman can be chosen from the remaining 7 persons in 7 ways. Therefore, total number of ways = = 56 ways.
Explanation
This question is similar to the example in the textbook context where Chairman and Vice-Chairman are chosen from 12 persons using permutation formula. Since the positions are distinct (order matters), we use permutation. The formula nPr = n!/(n-r)! gives us the number of ways to arrange r objects from n objects. Here n=8 and r=2, so 8P2 = 8!/6! = = 56.
Solution Steps
Step 1: Identify that this is a permutation problem since Chairman and Vice-Chairman are distinct positions (order matters).
Step 2: Apply the permutation formula: ⁸P₂ = 8!/(8-2)! = 8!/6!
Step 3: Calculate: 8!/6! = = 56 ways.