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Bunuel

2. How many positive integers of three different digits, each less than 400 can be formed from the digits 1, 2, 3, 4, 5 and 6?

# of 3-digit numbers possible to form with 6 digits 1, 2, 3, 4, 5, and 6 is \(P^3_6=120\). As number must be less than 400 then it should start with 1, 2, or 3 (with 3 out of 6 possible) --> so every 3 out of 6 number from 240 will fit --> [highlight]\(\frac{3}{6}*240=180\)[/highlight].



Bunuel, for the solution to the second question. Is there a typo? Should it not be \((3/6)*120\).
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saxenashobhit
A)
Number of digits that can take different place
3 5 4 3
_ _ _ _

3*5*4*3 = 180 numbers

B

3 5 4
_ _ _
Number below 400 = 3*5*4 = 60
thanx for solving!!
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Thanks Bunuel. Silly mistake on my part. Digits 3,4,5 and 6 will result in number greater than 3000. So 4, I agree.
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1)

the required # is > 3000
implies, The 1000s place shud be taken by 3,4,5,6, hence 4 posibilities. The remaining 3 digits can be selected from the remaining 5 available #s (note that one digit 3/4/5/6 is already selected for 1000s digit). Hence 5C3 ways, and those 3 #s can be arranges in 3! ways.

Hence answer is 4*5C3*3! = 240

2)

For the # to be < 400 the 100s digit shud be < 4 => one out of 3 choices 1/2/3

if the 100s digit is 1 then 10s digit can be selected in 5C1 and the units digit can be selected in 4C1 ways ==> 5C1*4C1=20
the same is the case with 2 and 3 too.

hence answer is 20+20+20 = 60
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