# Permutation and Combination

• Jan 17th 2011, 05:25 AM
somethingsmells
Permutation and Combination
Hello all, would very much appreciate some help with this question:

A code consists of blocks of 10 digits, 4 are zeroes and 6 are ones. Calculate the number of such blocks in which the first and last digits are the same as each other.

My final working is (8!/(6! x 2!)) + (8!/(4! x 4!)) = 98. I had two cases, the first case was when it started and ended with "0" and the second case where it started and ended with "1". The working is basically the arrangement of the remaining 8 digits divided by the nondistinct numbers. Could someone please tell me if I did it the right way?

My second question is why is this working wrong?

(4C2)(8!/(6! x 2!))+(6C2)(8!/(4! x 4!))

What I can't get out of my head is why do we not need to include any the selection of the two "1"s and "0"s at the start and end. I was wondering if for example, if you were selecting the "1"s that will be at the start and at the end, because all 6 "1"s are identical, and thus there is only 1 choice?

Thank you!
• Jan 17th 2011, 05:34 AM
Plato
Quote:

Originally Posted by somethingsmells
A code consists of blocks of 10 digits, 4 are zeroes and 6 are ones. Calculate the number of such blocks in which the first and last digits are the same as each other.

My final working is (8!/(6! x 2!)) + (8!/(4! x 4!)) = 98. I had two cases, the first case was when it started and ended with "0" and the second case where it started and ended with "1". The working is basically the arrangement of the remaining 8 digits divided by the nondistinct numbers. Could someone please tell me if I did it the right way?

My second question is why is this working wrong?
(4C2)(8!/(6! x 2!))+(6C2)(8!/(4! x 4!))

The part in blue is correct.

The part in red is not correct.
Because the ones are identical and zeros are identical.
• Jan 17th 2011, 05:35 AM
emakarov
You are right on both counts.

The answer is 98. We are counting the number of blocks of 10 numbers. If you select a pair of 1's to be at the ends and then select "another" pair of 1's, you still get exactly the same block of numbers; it does not have to be counted twice.