Mo2men wrote:Jay@ManhattanReview wrote:
Hi Mo2men,
Ways of choosing three 6s and two other and 6s = 1*1*1*9*9 = 81.
Let's arrange the three 6s. They can be arranged in 5C3 = 5C2 = (5*4)/(1*2) = 10 ways.
Total number of ways = 81*10 = 810. This is our numerator for the probability.
Hi Jay,
Thanks for your help.
As mentioned above, you calculated the way and then arrangement, could we solve it using permutation? Permutation cares about ways and places.
Thanks
Hi Mo2men,
Yes, we can do it, but that's a labored approach and not advisable.
For your interest, we can do this way.
The number of ways three 6s and two other than 6 digits can be arranged in 5P5 / 3! ways = 5! / 3! = 20 ways.
We know that if there are a few indistinguishable objects, we must divide the total number of ways by the factorial value of their repetition.
Here, there are three 6s, so we divided 5P5 by 3!.
However, 5! / 3! = 20 ways is still not what we desire. The 20 ways would include those passwords that have two 0s, two 1s, two 2s, two 3s, two 4s, two 5s,
No 6s, two 7s, two 8s, two 9s, when swapped at their fixed positions, though it does not give us new arrangement, give new ways; so we must get rid of these. Remember that a password such as 666
00 is wanted, but a password such as 666
00 is to be excluded as it is the same as the former. That's the reason I applied combination in my solution. The combination does the selection and takes care of repetitions.
Number of dummy ways = 5! / (3!*2!) = 10 ways. I divided 5! / 3! by '2!' to find out the repetitions.
So, the number of ways three 6s and two other than 6 digits can be arranged in = 20 - 10 = 10 ways.
Thus, total number of such passwords = 81*10 = 810. It's not advisable to do this way. It's unduly complicated.
Hope this makes sense.
