A package contains 13 resistors, 5 of which are defective. If 7 are selected, find the probability of gettin

Probability statistics question

Let D = defective
let N = not defective

DDDDDNN = (5/13)(4/12)(3/11)(2/10)(1/9)(8/8)(7/7) = 1 / 1287
But this event can happen 7c5 ways

7c5 ways = 21

21 / 1287 = 7 / 429
or
0.01631701631701631701631701631702
1.63% probability of getting exactly 5 defective resisitors

Number of possibel ways fo choosing 7 resistors out of 13 is
13C7=13!(7!6!)
=(8x9x10x11x12x13)/(1x2x3x4x5x6)
=1716

Number of possible ways of choosing 5 defectives from the 5 defectives, and selecting two non defectives out of the 8 non defectives is
(5C5)(8C2)=[5!/(5!0!)]*[8!/(2!6!)]
=7x8/(1x2)=28

So required probability=28/1716=0.0163

Total number of ways in which 7 resistors can be selected = 13C7

Number of ways in which 7 resistors can be selected such that 5 are defective
= 5C5 * 8C2
Probability = 5C5 * 8C2/13C7
= 7/429

getting 5 defective resistors out of 7 selected= 5C5. 8C2= 7x4=28ways
no of ways of selecting 7 out of 13= 13C7= 1716ways
probability= 28/1716