A key ring has 7 keys. How many different ways can the keys be

Think of choosing one specific key and placing it on the key ring. The remaining 6 keys can now be arranged in 6! ways. Once you have placed one key, you created a row (straighten the circle), so the regular rule for permutations apply.
It doesn't matter if you slide the keys clockwise or counter clockwise, they have the same relative position each to the other. For example, just for 3 keys, ABC and ACB are different arrangements. Mirror image arrangements are different, because you don't flip your key ring. At least I think this is what is assumed.

In the case of arrangements around a round table, it is obvious. For sure, you cannot flip your table and the people with it.
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Thanks Eva. I think that we need to apply circular permutations for necklace. Isn't it? https://www.transtutors.com/math-homewor ... tions.aspx

I read this concept while preparing for Indian Engineering entrance examination.... not sure....

Necklace, with spherical beads, can be flipped, so symmetrical arrangements can be considered identical.
But, if not explicitly stated, I think you shouldn't assume anything like this.
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if you are like me who chosed 7! [as is evident 42% of us chose this] then please refer to below link for an awesome explanation

https://www.youtube.com/watch?v=tn6sc0q2ILg

The thing about a key ring is that, since the keys can spin around the ring, the positions of the keys are all relative to each other.
So, let's take the task of arranging the keys and break it into stages.

Let's call the 7 keys A, B, C, D, E, F and G

Stage 1: Place the A key on the ring.
We don't care about the "position" of the ring because all positions are the same. In other words, if I asked you "In how many ways can we place 1 key on a key ring?" the answer is 1.
So, we can complete stage 1 in 1 way.

Stage 2: Select a key to go to the immediate right of key A (i.e, clockwise from key A)
There are 6 remaining keys to digits from which to choose, so we can complete this stage in 6 ways.

Stage 3: Select a key to go to the immediate right of the key we selected in stage 2.
There are 5 keys remaining, so we can complete this stage in 5 ways.

Stage 4: Select a key to go to the immediate right of the key we selected in stage 3.
There are 4 keys remaining, so we can complete this stage in 4 ways.

Stage 5: Select a key to go to the immediate right of the key we selected in stage 4.
There are 3 keys remaining, so we can complete this stage in 3 ways.

Stage 6: Select a key to go to the immediate right of the key we selected in stage 5.
There are 2 keys remaining, so we can complete this stage in 2 ways.

Stage 7: Select a key to go to the immediate right of the key we selected in stage 6.
There is 1 key remaining, so we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 7 stages (and thus arrange all 7 keys) in (1)(6)(5)(4)(3)(2)(1) ways (= 720 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT, so be sure to learn this technique.

RELATED VIDEOS FROM OUR COURSE
Fundamental Counting Principle (FCP)



Fundamental Counting Principle - example

Arranging keys in a ring means arranging keys in a circular pattern

so arranging 7 keys can be done in (7-1)! = 6! = 720

Option D

The circular arrangements of n distinct objects is represented by (n-1)! Because one of the n objects needs to be fixed.

Out of 7 different keys one needs to be fixed and remaining 6 keys can be arranged in 6! Ways
Hence, 6!=720 ways

Answer: Option D

GMATinsight : There is a minor mistake (see highlighted part). Can you please update it to D.

Thank you!!!

Modified it.

When reading this problem, we must notice that we are organizing the keys around a ring, in other words, a circle. Since we are arranging items around a circle, 7 keys can be arranged in (7-1)! = 6! = 720 ways.

Answer: D
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Official solution from Veritas Prep.

In a permutations problem that involves a circular arrangement (such as keys on a ring, or people seated around a circular table), the number of permutations is equal to (N - 1)!. This is because there is no beginning or end to the ring (or no left end or right end). Consider the three-item arrangement of A, B, C. In a row, ABC is different from BCA, but if they're in a circle they're the same arrangement: B is between A and C, A is between C and B, etc.

So in this problem, with 7 keys, the calculation is (7 - 1)!, which is 6! which equals 720.
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