akadmin wrote:

Why have you done 5C3 and not 5!/3!. In my opinion the order matters so the 5!/3! should remove the three 2s arrangeemnt out of the equation

chetan2u wrote:

hi i think it does not req such long calculations my ans A...

there are 5 digits each place can have any of ten digits so total posb=10^5...

now three places only two can be there and rest two can have any of remaining nine digits...

so posb=1*1*1*9*9*5c3..

prob=.81%... although i might be still missing something.. ill try again

Responding to a pm:

**Quote:**

I am confused on the arrangement part of it. 3 2s and 2 not2s cannot result in 5C3 or 5!/2!3! . My answer is 5!/3! since i am considering this as a permutation of 5 elements with 3 identical and other 2 may or may be not identical. How do I solve this question with arrangement with repetition standpoint.

5C3 means you are selecting 3 out of 5.

Here, you are selecting 3 spots out of the available 5 spots (1st digit, 2nd digit, 3rd digit, 4th digit, 5th digit). Now you have all such arrangements:

____ 2 ____ 2 2

2 ____ ____ 22

2 ____ 2 ____ 2

... etc

In the two blanks, you can have any of them 9 remaining digits. You will get all ids of the from

23224

23223

32522

etc

i.e. all ids where the remaining two digits are the same or are distinct.

So favourable cases = 5C3 * 9 * 9

Your Method: In case, you want to specifically arrange, you will need to take two cases:

Case 1: Three 2s and two distinct other digits

No of ways = 9C2 * 5!/3! = 720

Case 2: Three 2s and two another digit

No of ways = 9C1 * 5!/3!*2! = 90

Total no of favourable ways = 720 + 90 = 810

Required probability = 810/100000 = .81%

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Karishma

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