John has 5 friends who want to ride in his new car that can accommodat

Question

John has 5 friends who want to ride in his new car that can accommodate only 3 passengers at a time. How many different combinations of 3 passengers can be formed from the 5 friends?

A. 3
B. 8
C. 10
D. 15
E. 20

abhimahna · Answer

5 Friends, We have to select two friends out of these 5(as the Car can accommodate only 3 passengers at a time, John + other 2),

So, 5C2 = 10

Hence, C

tanmay056 · Answer

I am not sure if - 5C2 is going to be correct one .... It seems to me that 5C3 is going to be more precise answer. Because we need to select group of 3 people out of 5. Though both of them will result in 10 . Please clarify of choosing 5C2 over 5C3.

Abhishek009 · Answer

5C_3 = 10

Answer will be (C) 10

vijaisingh2001 · Answer

both 5c3 and 5c2 are same= 10

passenger includes the driver or not that is the big question here? I believe that is true based on the advertisements of SUV - example 07 seaters, means the driver is included here

so the answer should be 5c2
in case in future these numbers get change

chelsiecan · Answer

Use equation nCr =  \frac{(n!)}{ }  to choose n objects taken r at a time.

5C3 =  \frac{(5!)}{(2!3!)}  = 5 x 2 = 10

Answer C. 10

broall · Answer

$5C2 = 5C3$

Note that $C^k_n =\frac{n!}{k!(n-k)!}$

$C^{n-k}_n=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{(n-k)!k!}$

Hence $C^k_n = C^{n-k}_n$

That's why $C^3_5 = C^2_5$

Now let's compare $5C2$ vs $5C3$.

If you choose 3 out of 5, there are $5C3$ different ways.

If you choose 3 out of 5, that means you drop 2 of 5 and 3 left. Thus there are $5C2$ different ways.

Two different ways with different concept, but have the same result. So you could choose any one which suits you.

akoundourakis · Answer

I've seen in other questions in which I would have to multiply Combinations or subtract an integer from the Combination. Can you tell me why here we only do 1 Combination?
I think when you need to multiply is when there are, for example, women and men. And to subtract I'm not sure of an instance at the moment.

atr0038 · Answer

Does it really take people only 22 seconds on average to get this question correct? I'm lagging behind, because I took almost double that amount of time to read the question, process it, and do then math. It took me like 38 seconds to solve this, but how can I get as fast as the average person at solving these types of problems? There must be some kind of shortcut if that many people can solve it that quickly.

nust2017 · Answer

Can anyone please explain the solution to this question for someone with average quant skills. I took more than 1 minute to answer the question, and the average time taken for the question is 22 seconds :O

BLTN · Answer

Dear nust2017 try to think in this way, there are only 3 places: + + + But 5 friends that signifies 2 friends without places: X X Hence we can represent: + + + X X That is 3! and 2! our denominator Total 5 friends or 5! is a numerator \frac{5!}{3!*2!} = 10 Hope it helps

rxb266 · Answer

The different combinations of 3 passengers that can be formed from the 5 friends:

5C3 = 5! / (3!x2!) = 10
 
option C

evagmat · Answer

Hope this draft helps!

Rahulkunar5657 · Answer

We need to find the number of ways to choose 3 passengers from 5 friends. This is a combination problem, where order does not matter. The formula for combinations is:

nCr = n! / r!(n−r)!

Here, n = 5 (the total number of friends) and n = 3 (the number of passengers to be chosen). 
Plugging in these values, we get:
5C3 = 5! / 3!(5−3)! = =10

So, the number of different combinations of 3 passengers that can be formed from the 5 friends is 10. 
The correct answer is C.

ManifestDreamMBA · Answer

Need to find combinations of 3 passengers that can be formed from the 5 friends
5C2 = 10

Answer is C

John has 5 friends who want to ride in his new car that can accommodat

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John has 5 friends who want to ride in his new car that can accommodat

Prep Toolkit

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