Question

A certain system has two components. There are 6 different models of the first component and 10 different models of the second. Any first component can be paired with any second component. A salesman must select 2 of the first component and 3 of the second to take on a sales call. How many different sets of components can the salesman take?

Answer 1

Answer: 1800

Step-by-step explanation:

Given : A certain system has two components.

Number of different models of the first component = 6

Number of different models of the second component  = 10

A salesman must select 2 of the first component and 3 of the second to take on a sales call , so we use combinations ( ∵ order of selection not matters)

The number of different sets of components can the salesman take = $^{6}C_2\times^{10}C_3$

$\dfrac{6!}{2!(6-2)!}\times\dfrac{10!}{3!(10-3)!}\ \ [\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}]$    

$=1800$

Hence, the number of different sets of components can the salesman take = 1800