One of the approaches:
Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 4 to send only one "representatives" to the committee: 4C3=4.
But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.
Total # of ways: 4C3*2^3=32.
hope it helped a bit
The average number of failures can be found by multiplying .41 (equivalent to 41%) by 375. When you do this, you get 153.75 failures. Rounded to the nearest 10th is 153.8 failures; the nearest whole number, 154 failures.
