GRE counting problem

Question

I have approached this question the following way.

For Quantity $A$: The number of ways to pick $3$ cards including $1$ is : $1\cdot 4\cdot 3=12$ ways. [$1$ is fixed]

For Quantity $B$: The number of ways to pick $3$ cards excluding $1$ is : $4\cdot 3\cdot 2=24$ ways resulting $24$ which means $B$ is a greater quantity, but I got it wrong.

Can someone explain to me where I went wrong? I did not find Manhattan's explanation to this problem sufficient, they use some weird shortcuts. Can someone break this down if possible?

Answer 1

They’re counting $3$-card sets, not sequences of $3$ cards. There are $\binom42=6$ pairs of cards that could be combined with the $1$, so Quantity A is $6$. There are $\binom43=4$ ways to choose $3$ of the non-$1$ cards, so Quantity B is only $4$.

Since the numbers are so small, here are the $3$-card hands that include the $1$:

$$\{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,3,4\},\{1,3,5\},\{1,4,5\}$$

And here are the $3$-card hands that don’t include it:

$$\{2,3,4\},\{2,3,5\},\{2,4,5\},\{3,4,5\}$$

Answer 2

What you have counted is the number of permutations with condition $A$ and condition $B$. This is correct if the problem had stated that the order was significant. Usually for problems like this it is assumed that the order does not matter. The reason for this is because if you pick card $4$, card $3$, and card $1$, then this is the same as picking card $1$, card $3$, and card $4$. No matter what way you pick these three cards you will always have the same hand. So this problem is a combination problem. Brian M. Scott has written out all of the combinations with the given conditions so that you can see it better.

Answer 3

Its not that complicated to solve this question! For column A, Since we have to pick card #1 in the first draw, then the number of ways of doing that is only 1 way. The next 2 draws don't matter. So, {1,2,3} is same as {1,3,2}. So, order doesn't matter. Hence, this is a combination question. Therefore, total ways of choosing 4 cards in 2 draws is 4C2 ways = 6 ways. Therefore total ways = 1way x 6ways = 6 ways

Now, for Column B. To calc. excluding card 1, we'll need all scenarios where we don't pick card 1. Which is same as "Total ways - Total ways of choosing card 1 at least once" We have totaly ways of choosing card 1 as 6 from before. Total ways without any restrictions is 5C3 ways = 10 ways Therefore, we get 10-6 = 4ways. Hence, Column A is greater. Answer

Answer 4

As for this solution : "For Quantity A: The number of ways to pick 3 cards including 1 is : 1⋅4⋅3=12 ways. [1 is fixed]

For Quantity B: The number of ways to pick 3 cards excluding 1 is : 4⋅3⋅2=24 ways resulting 24 which means B is a greater quantity, but I got it wrong.

Can someone explain to me where I went wrong? I did not find Manhattan's explanation to this problem sufficient, they use some weird shortcuts. Can someone break this down if possible?"

The main error that you're making is you're considering certain scenarios twice! For eg., consider that you have chosen card 1 in first draw. Now, you draw card 4 and card 3 in the second draw. This will be same as drawing card 3 and card 4. Hence, you need to eliminate the duplicate cases. That's why you're getting more ways in the for column A.

As for column B, that is not the right approach to solve the question. Although you're considering that you will draw the other 4 cards first, then the other 3 cards and then the other 2 cards. But, the problem with this solution is that you're completely ignoring the fact that card 1 is still in the deck! So, you cannot guarantee that the next card you pick is not going to be 1 because you just dont know what the next card is going to be! Moreover, you're considering extra cases like first case. For eg., if you draw {2,3,4} or you draw {2,4,3} both ways are the exact same. At the end you have cards 2,3 and 4. Hence, there are actually NOT 24 ways. If you want to look at it in layman's terms, then drawing 2,3,4 is same as 2,4,3 or 3,2,4 or 4,3,2 etc. and 2,3,4 is same as 2,3,4 etc. and 2,3,5 is same as 2,5,3 etc. and finally 3,4,5 is same as 5,4,3 etc. Thus we see there are actually only 4 UNIQUE ways :)