Talk show host Ralph Burke has exactly one guest on his show each day, and Burke's show airs every Monday to Friday...

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Princeton Review

Talk show host Ralph Burke has exactly one guest on his show each day, and Burke’s show airs every.Monday through Friday. Burke always schedules politicians on Mondays and Wednesdays, actors on Tuesdays and athletes on Thursdays, but can have a guest of any one of these three kinds on Friday. No guest appears more than once per week on Burke’s show. If Burke has five politicians, three actors and six athletes he could invite, and if no politician is also an actor or an athlete and no actor is also an athlete, how many different schedules of guests from Monday to Friday could Burke create?

A. 30
B. 1,200
C. 3,600
D. 4,500
E. 6,300

OA C

[deloitte247]

Given that:
- Ralph hosts one guest per day
- The show runs for Mondays to Fridays (5 days)
- Politicians are scheduled for Mondays and Wednesdays (2 days)
- Actors come up on Tuesdays (1 day)
- Athletes come up on Thursday (1 day)
- Any of politician, actor or athlete can come up on Friday
- No guest appears more than once per week
- Ralph has 5 politicians, 3 actors, and 6 athletes

=> How many different schedules of guests from Monday to Friday could Burke create?

$$5\ politicians\ can\ be\ selected\ on\ Mondays\ and\ Wednesdays$$
$$in\ 5C1=\frac{5!}{1!\left(5-1\right)!}=5\ ways\ on\ monday$$
$$in\ 4C1=\frac{4!}{1!\left(4-1\right)!}=4\ ways\ on\ wednesday$$
$$-\ 3\ actors\ can\ be\ selected\ on\ tuesday\ in;$$
$$3C1=\frac{3!}{1!\left(3-1\right)!}=3\ ways\ on\ tuesday$$
$$-\ 6\ athletes\ on\ thursday\ can\ be\ selected\ in;$$
$$6C1=\frac{6!}{1!\left(6-1\right)!}=6\ ways\ on\ thursday$$


If guests have been selected between Monday and Thursday, the remaining guests to select from = 5 + 3 + 6 - 4 = 10
$$10C1=\frac{10!}{1!\left(10-1\right)!}=10\ ways$$
Total number of possible schedules that can be created = 5 * 4 * 3 * 6 * 10 = 3600

Answer = C