Answer:
120
Explanation:
To solve this question we would be using combination formula.
The formula for combination is given as:
C(n, r) = nCr
nCr = n!/r! ×(n - r)!
In the above question,
n = 10 men
r = 3 operators
Hence,
nCr = n!/r! × (n - r)!
10C3 = 10! /3! × (10 - 3)!
10C3 = 10!/3! × 7!
10C3 = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 /(3 × 2× 1 ) ×(7 × 6 × 5 × 4 × 3 × 2 × 1)
10C3 = 10 × 9 × 8 / 3 × 2× 1
10C3 = 720/6
10C3 = 120
Therefore, the number of groups of 3 operators that are possible is 120.