PS: Using anagrams Correctly

[GK]

A shipment of 10 light bulbs includes 3 that are defective. In how many ways can a customer purchase 4 of these bulbs and receive at least two of the defective bulbs?

I don't think that I'm using MGMAT's anagram method correctly:

Ways to purchase 4 bulbs with at least 2 defective: (D = defective, OK = working)
D D OK OK = 4!/(2!*2!) = 6
D D D OK = 4!/(3!*1!) = 4

Total = 6 + 4 = 10 ways.

Can you please tell me what I'm doing wrong.

The other solution (correct one) for the problem is as follows:

3C2*7C2 + 3C3*7C1 [choosing 2 defective out of 3 defective and choosing 2 ok out of 7 ok OR choosing 3 defective out of 3 defective and choosing 1 ok out of 7 ok].

What is the best way to approach this kind of problem? In what case would you use the anagaram method (the former) vs. the latter?

Thanks a lot.

Source: My own question

[StaceyKoprince]

The way you're using the method only accounts for 4 bulbs total - it's as though you only start with 4 in the first place. But you really start with 10, of two different types (defective or not). You have to make sure you factor that in.

The two ways to meet the requirements (4 bulbs, at least 2 of which are defective) are to have 2 okay and 2 defective, or 1 okay and 3 defective.

Option one: 2 okay and 2 defective. Split into two parts: find the number for each part (okay, defective) and then multiply them (because we want both - "and"). First part, okay bulbs:
A B C D E F G (my 7 okay bulbs)
Y Y N N N N N (that is, I choose 2 and don't choose 5)
The above gives me 7!/(2!5!)
Second part, defective bulbs:
A B C (my 3 defective bulbs)
Y Y N (that is, I choose 2 and don't choose 1)
The above gives me 3!/(2!1!) or simply 3!/2!
Multiply these two together for the total options here: [7!/(2!5!)]*[3!/2!] - there's the number for my first option.

Option two: 1 okay and 2 defective. Again, split into two parts. First part, the above process gives me 7!/(1!6!) or simply 7!/6! Second part, the above process gives me 3!/3! or simply 1. Multiply these together: 7!/6!*1 or simply 7!/6!.

Now I want either option one or option two, so I add those two together: [7!/(2!5!)]*[3!/2!] + 7!/6!. And this matches what you have for the "official" math way - same answer.

[Guest]

Shouldn't there be a negative sign between (n and k) in the denominator if it's a combination? I'm confused - the explanation is very confusing! Sorry!

[StaceyKoprince]

The explanation given above specifically uses the ManhattanGMAT anagram method to solve the problem. If you haven't taken our course or studied from our books, then this approach won't make much sense to you. Please let me know whether you are familiar with this method. Please also clarify where you think a negative sign should be - I don't use the variables n and k in my explanation.