=>

X = { 3, 6, 9, 12, 15, 18, 21, ... } and Y = { 7, 14, 21, ... }
The following pairs ( x, y ) satisfy x + y <= 21:
(3, 7), (3, 14), (6, 7), (6, 14), (9,7) and (12,7).
There are 6 such pairs, (x,y).

Therefore, B is the answer.
Answer: B

Set X=> contains positive integers which are multiples of 3.
$$X=\left\{3,6,9,12,15,18,21,24,...,\ \inf inity\right\}$$
Set Y=> contains positive integers which are multiples of 7
$$Y=\left\{7,14,21,28,35,\ ...\ ,\ \inf inity\right\}$$
$$Define\ a\ new\ set\ of\ X+Y\ with\ constructor\ x+y\ where\ x\ is\ an\ element\ of\ set\ X\ and\ Y\ is\ an\ element\ of\ set\ Y\ which\ must\ be\ \le21.$$
$$X+Y=\left\{\left(7+3\right),\left(7+6\right),\left(7+9\right),\left(7+12\right),\left(14+6\right),\left(14+3\right)\right\}$$
$$X+Y=\left\{10,13,16,19,20,17\right\}$$
$$Total\ elements\ of\ X+Y=6\ elements\ \le21$$

Answer is option B.