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A set of cards is numbered $1$ through $6.$

• Quantity A: The number of ways to pick $3$ of the $6$ cards such that card number $1$ is included.
• Quantity B: The number of ways to pick $3$ of the $6$ cards such that card number $1$ is excluded.
1. Quantity A is greater than Quantity B.
2. Quantity B is greater than Quantity A.
3. The two quantities are equal.
4. The relationship cannot be determined from the information given.

They’re counting $3$-card sets, not sequences of $3$ cards.
There are $^{5}\text{C}_{2} = 10$ pairs of cards that could be combined with the $1,$ so Quantity A is $10.$
There are $^{5}\text{C}_{3} = 10$ ways to choose $3$ of the non-$1$ cards, so Quantity B is only $10.$
So, the answer is Option C.

NOTE that for a set of cards is numbered $1$ through $5,$ the following will happen:
There are $^{4}\text{C}_{2} = 6$ pairs of cards that could be combined with the $1,$ so Quantity A is $6.$ There are $^{4}\text{C}_{3} = 4$ 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:

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

NOTE that for a set of cards is numbered $1$ through $n,$ where $n \geq 7,$ the Quantity B will be higher.

The typo has been corrected now.

NOTE that for a set of cards numbered 1 through n, where n≥7, the Quantity B will be higher.

Verification of the above statement:

Quantity A: The number of ways to pick 3 of the 7 cards such that card number 1 is included.

6C2 = 15 pairs will combine with card 1 which is included already

Quantity B: The number of ways to pick 3 of the 7 cards such that card number 1 is excluded.

6C3 = 20 ways to select 3 cards from 6 cards as 1 is excluded.

Hence, Quantity B is higher.

set of cards is the main keyword for this question.

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