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by MBA.Aspirant » Mon Nov 07, 2011 3:29 pm
A diner serves a lunch special,consisting of soup or salad,a sandwich,coffee or tea & a dessert. If the menu lists 3 soups,2 salads,7 sandwitches ,& 8 desserts,how many different lunches can you choose??(note :-two lunches are different if they differ in any aspect.)

A>22
B>280
C>336
D>560
E>672
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by user123321 » Mon Nov 07, 2011 3:42 pm
IMO D

soup or salad combinations = 3+2 = 5
sandwiches = 7
tea or coffee = 2
desserts = 8

so using product rule, possible combinations are 5*7*2*8 = 560 ways

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by MBA.Aspirant » Mon Nov 07, 2011 3:55 pm
Thanks, what i don't get is the soup or salad part. I made it 3C1 * 2C1 and got E as a final result. Can you explain this part?
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by user123321 » Mon Nov 07, 2011 4:48 pm
the number of ways of picking one soup "or" one salad salad is 3C1+2C1 = 5 ways
what you did was the number of ways of picking one soup "and" one salad then it is 3C1*2C1 = 6 ways.

you just need to distinguish between and & or clause in problem statement and rest is easy.

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by MBA.Aspirant » Thu Nov 10, 2011 1:01 pm
Thanks for your help

(3+2) * 7C1 * 8C1 * (1+1) = 560
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