aarthis2
ezhilkumarank
u0422811
Company A is twice the size of its nearest competitor, Company B. Company A has 13 employees, 8 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift?
(a) 56
(b) 231
(c) 336
(d) 350
(e) 406
Looks easy: -- 8C4*5C1 + 8C5 -- 406 (E).
Could you please explain how you got 8c5.I used this logic...Totally 13 employees out of which 8 are union emp.In order to have atleast 4union employees in any shift.I splitted it as 8c4*5c1(selecting 4 emp from 8 union emp and 1 employee from remaining 5 non-union employees).I got the answer as 350.I didnot understand how you got 8c5.Kindly help me.
There are total of 13 employees, 8 of whom belong to the union and 5 doesn't. Out of 5 people working a shift
at least 4 must belong to the union.
Now, at lest 4 out of 5 means that
4 or all 5 employees must belong to the union:
Out of 5 people working a shift 4 employees belong to the union and 1 doesn't: \(C^4_8*C^1_5=350\);
Out of 5 people working a shift
all 5 employees belong to the union: \(C^5_8=56\);
Total # of ways: \(350+56=406\).
Answer: E.
Hope it's clear.