4 cards are drawn no replacements odds that atleast one is a club?

Question

ive done this a few times but the book says im wrong(i think it is but)

anyway i do 1- [(.75^4)/(52*51*50*49)]

Answer 1

chance of non club # 1 = 39/52 (39 non clubs out of 52 cards)

chance of non club #2= 38/51 (38 non clubs out of 51 cards)

chance of non club #3 = 37/50 (37 non clubs out of 50 cards)

chance of non club #4 = 36/49 (36 non clubs out of 49 cards)

1- (39 * 38 * 37 * 36) / (52 * 51 * 50 * 49)

So it's about 70% that if you pick four cards, at least one will be a club