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.