Probability Problem

[VadimC819]

There are 6 different colored neck ties (red, orange, yellow, green, blue and indigo) and six different colored shirts (red, orange, yellow, green, blue and indigo) that must be packaged into gift boxes. If each box can only fit one necktie and one shirt, what is the probability that ALL of the boxes will contain a necktie and a shirt of the same color.

Is the answer here 1/120 or 1/720 why?

When looking at the probability my logic is


Anagram Method 6!/5! + 6!/5! to get the number of all combos of matching clothing
(6!*2) = number of total combinations

This yields 1/120. Why is this logic incorrect?

[esledge]

When looking at the probability my logic is


Anagram Method 6!/5! + 6!/5! to get the number of all combos of matching clothing
(6!*2) = number of total combinations

This yields 1/120. Why is this logic incorrect?

The short answers are:
(1) You won't have a plus sign anywhere in the solution for this question, because plus is for "or" situations, such as "what is the probability that the red shirt is paired with the blue neck tie or with the yellow neck tie?"
(2) Your anagram method indicates 6+6 = 12 number of combos of matching clothing, but there is really just 1 way that can happen. Think about why, without math and just with real-life logic. Imagine the matching clothing pairs in their 6 respective boxes. If you switch the box that the red pair and the blue pair are in, you don't create a new combo--it's the same one. If you switch the blue neck tie with the red neck tie, now it's not a "winning" matching combo anymore.

There are 6 different colored neck ties (red, orange, yellow, green, blue and indigo) and six different colored shirts (red, orange, yellow, green, blue and indigo) that must be packaged into gift boxes. If each box can only fit one necktie and one shirt, what is the probability that ALL of the boxes will contain a necktie and a shirt of the same color.

Is the answer here 1/120 or 1/720 why?

The answer is 1/720 = (1/6!). Here's my logic:

Draw 6 boxes in a row on your paper, each with space for a shirt and a tie, which I will just put S and T placeholders for right now:
[ST][ST][ST][ST][ST][ST]

Pretend I am physically selecting the items and putting them in the boxes. For the first box, there are 6 shirts and 6 ties to choose from. When I move on to the second box, there are 5 shirts and 5 ties to choose from, and so on. "And" means multiply:
[6*6][5*5][4*4][3*3][2*2][1*1] = 6!6!

We don't divide by 2 within each box, because "order of the items in each box matters," so to speak: shirts are different from ties. However, imagine that we have completed the packing of the items in boxes, so we have particular/fixed pairings of shirts and ties. Imagine we tape the boxes shut so we don't change the shirt & tie pairings. Now we shuffle the boxes around, putting them in a different order left-to-right. The number above counts that arrangement as if it's different, but it really isn't: the "order of the boxes does NOT matter." So, divide the number above by 6! to correct for the fact that all 6! orders of the taped boxes are really just 1 scenario of shirt and tie pairings.

Number of ways "ALL of the boxes will contain a necktie and a shirt of the same color": just 1
Total number of unique necktie and a shirt combinations: 6!6!/6! = 6! = 720

Answer: winning combos/ total combos = 1/720