2013-10-24, 05:36 AM
In (a) you can look at it as picking the 'last 7' which aren't cordless first: 14!/7! ways to take those from the other two groups, and the remaining 14 can be sorted 14! ways. Then divide that by 21! which is the total number of permutations. That 7 of them are identical doesn't matter here. 14!^2 / (21!*7!) gives 0.029514 agreeing with shouri.
My tendency is to not use npr/ncr and just do straight up factorials cause it's easier for me to not screw up; 14!/7! is obviously the same thing as 14p7. When you have (7!*14p7)/(21p14) it expands to (7!*14!/7!)/(21!/14!) = 14!^2/21!, which is off by that one extra factor of 7!. 14p7/21p14 = 0.0295 so I'm not sure what issue you had that made it come out 0. Some calculators don't handle factorials well; I used Wolfram Alpha.
My tendency is to not use npr/ncr and just do straight up factorials cause it's easier for me to not screw up; 14!/7! is obviously the same thing as 14p7. When you have (7!*14p7)/(21p14) it expands to (7!*14!/7!)/(21!/14!) = 14!^2/21!, which is off by that one extra factor of 7!. 14p7/21p14 = 0.0295 so I'm not sure what issue you had that made it come out 0. Some calculators don't handle factorials well; I used Wolfram Alpha.