bhavinshah5685
we units 1,2,3,4,5,6,7,8,9
total 9 digits
Now we want arragement xxyy
x - can be selected in 9 ways
x - same as first x - 1 way
y - can be selected from remaining 8 digits - 8 ways
y - same as first y - 1 way
so total# of ways = 9*1*8*1 = 72
not the 4 digits can be organised in 4C2 ways = 6 ways
So total arragements = 72*6 = 432
Am I wrong in thinking? & I didnt understand the concept mentioned by Bunuel here...
P & C is my weakest point in Quant

the 4 digits can be organised in 4C2 ways = 6 waysYou have to divide by 2, so only 3 ways.
Or, 9 possibilities for the first digit, this can be placed for the second time in 3 places, then in the remaining places have another different digit - 8 possibilities.
A total of 9*3*8=216.
4C2 would work if you choose two distinct digit out of 9, which is 9C2 = 36 and you don't care about the order in which you have chosen them, take 2 of each type, then arrange them, which is 4C2 = 6.
This will give you again 9C2*4C2 = 36*6 = 216.
You cannot in one stage take order into account, then in the next stage ignore order. Choosing two digits as 9*8 means you distinguish between which one is chosen first. For example, once you considered x = 4 and y = 1, which you can arrange in 4C2 = 6 ways. But in this 6 arrangements are also include those with a 1 in the first place.
Then you considered the pair x = 1 and y = 4, for which again 4C2 = 6 arrangements, but these are identical to the previous ones. 4C2 counts arrangements without distinguishing whether you first place x or y.