There were 6 members in Mr. George's family: his wife, 2 daughters, on

Question

There were 6 members in Mr. George's family: his wife, 2 daughters, one son and mother and himself. They had 6 cars out of which 2 were identical cars of brand A, one car of brand B, third was of brand C, one was of brand D& the sixth car was brand of E. If all family members want to use separate cars, in how many ways can they choose if Mr. George's son refuged to take the car of brand A?

A. 180
B. 200
C. 240
D. 360
E. 480

A. 180
B. 200
C. 240
D. 360
E. 480

Sabbir266 · Answer

The son has 4 options to choose from & remaining members can choose 5!= 120 ways.

So, total number of ways = 4×120=480

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Regor60 · Answer

Remember that you're doing a permutation, which assumes order matters, but does order matter with the 2 people assigned car A ?

pleasefix · Answer

Good observation, but with the current phrasing of the question it seems to be assuming it is two separate cars.

If it was phrased more like; "(...) in how many ways can they choose a brand if Mr. George's son refused to take the car of brand A ".

In that case it is basically the same, but now the family has only 5 cars and 4 brands.
To me, that sounds like the ordering should be; 5!/2!=60 options. Since the son has 4, the total ways to combine would be 240.

notgivingupami · Answer

For the same stem, if the question had the constraint that the Son refused to take car of Brand A, then would the options for the son be = 4!/2! ( since there are 2 cars of Brand A)

martinfzb · Answer

the complete equation is 4C1 * 5!/2!=4*120/2=240
Note: divided by 2! is to remove the duplicated on 2 identical cars of brand A

samarpan.g28 · Answer

The son can choose a car in 4C1 way. Now the remaining 5 cars are chosen in 5! ways. So total arrangement 4C1*5!/2!=(4*120)/2=240(C).

sachi-in · Answer

Equivalent to giving 6 students 6 awards ( where each car is a award with two being identical )

One such allocation can be:

A B C D A E
George Wife Mother Son Daughter1 Daughter2

Both brand A cars are identical ( in real life cars of same brand are not identical but this problem states car brands to ensure that cars can only be identified by brands not number plates )

Hence 6!/2! ways to allocate cars.

Now Son wont drive Brand A. Considering Son drives Brand A we have 5! ways to allocate the rest of the brands.
Hence with Son driving A there are 5! allocations.

6!/2! - 5! = 240

Oppenheimer1945 · Answer

G, G', S, D1, D2, M chooses A,A,B,C,D,E

S doesno't choose A, Hence he has 4 choices.

rest 5 can select b/w AACDE, 5!/2!

No of ways= 4*60=240

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