In your office 4 posts have fallen vacant. In how many ways a selectio...
4 posts,31 candidates .out of which 1 always included..from the 4 positions 1 position filled rest 3 positions also from 31 candidates 1 always included rest 30 candidates therefore we can arrange 30 candidates in 3 positions so,30c3
In your office 4 posts have fallen vacant. In how many ways a selectio...
To solve this problem, we need to use the concept of combinations.
Given:
- There are 31 candidates in total.
- 4 posts have fallen vacant.
- One candidate is always included in the selection.
We need to find the number of ways a selection can be made out of the 31 candidates, considering that one candidate is always included.
Solution:
Step 1: Select the candidate who is always included
Since one candidate is always included, we have only one choice for this candidate.
Number of ways to select the candidate who is always included = 1
Step 2: Select the remaining candidates for the vacant posts
Out of the remaining candidates (31 - 1 = 30), we need to select candidates for the 4 vacant posts.
Number of ways to select the remaining candidates = 30C3
(30C3 represents selecting 3 candidates out of 30)
Step 3: Multiply the number of ways from Step 1 and Step 2
To find the total number of ways to make a selection, we need to multiply the number of ways from Step 1 and Step 2.
Total number of ways = 1 * 30C3
Simplifying the expression:
30C3 = (30!)/(3!(30-3)!)
= (30!)/(3!27!)
Cancelling out common terms in the numerator and denominator:
= (30 * 29 * 28)/(3 * 2 * 1)
= 4060
Therefore, the correct answer is option 'A', which is 30C3.