In how many ways can a group of 2 people from different fields be chosen if a company has 5 mathematicians,?

Question

6 electrical engineers, 2 teachers, and 1 physicists?

Answer 1

You can calculate all the possible pairings as a previous respondent suggested, but the more different groups there are, the harder this gets.

I find it easiest to take the total number of ways to select two people from the total of 14 (5 + 6 + 2 + 1), and subtract how many of those combinations are two people from the same group.

(For example if you had 10 different professions, you'd have to do 10*9/2 = 45 pairings as suggested by other respondents, but only 11 numbers for mine.)

14C2 = 14*13/2 sets of two people, is 91. This is the total number of pairings of two people whether or not they are from the same group.

Of those 91...
(a) 5C2 = 5*4/2 = 10 have two math,
(b) 6C2 = 6*5/2 = 15 have two EE,
(c) 2C2 = 2*1/2 = 1 has two teachers.
(There are no groups of two physicists since there's only one physicist.)

The remainder is:
91 - 15 - 10 - 1 = 65
... so there must be 65 groups of two people that do not have two of the same profession.

Answer 2

Imagine chosing a mathematician and an engineer. 60 ways. Right?
Now, what other combinations of professions are there...
m&t, m&p, e&t, e&p, t&p. Right!
now chose from each of these pairings,

add em all up.
you done it!

Answer 3

Combinations:
M-E: 30
M-T: 10
M-P: 5
E-T: 12
E-P: 6
T-P: 2

Sum: 65