Math Question

i) How many different samples can be chosen?
40C4 = [(40)(39)(38)(37)]/[(4)(3)(2)(1)] = 91,390
ii) How many samples will contain at least one defective board?
The number of samples that contain NO defective board
= 37C4 = [(37)(36)(35)(34)]/[(4)(3)(2)(1)] = 66,045
Samples contain at least one defective board = 91,390 - 66,045 = 25,345

i) 0 are not together
Consider that the six '1' queue up leaving seven spaces in between each of them:
( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( )
Now the four '0' select the spaces to be placed then form this 10-digit code.
# of Ways = 7C4 = 35
ii) first and second digits are the same
Scenario 1 : First and second are '0'
8!/(2!)(6!) = 28
Scenario 2 : First and second are '1'
8!/(4!)(4!) = 70
Total # = 28 + 70 = 98

i)40C3
=9880
ii)3*37*36+3*37+1
=4108
i)35
ii)8*7/2+8C4
=28+70
=98