Permutation

With the digits given the numbers that can be made can only be three digits.

The first place can be filled only by 4 or 5 or 6 = 3 ways.
The second place can be filled by any of the 6 digits = 6 ways.
The third place can be filled by any of the 5 remaining digits + the digit used in second place above = 6 ways
Therefore total number of ways = 3 x 6 x 6 =108 ways(numbers)

Hey thanks for the reply..

I agree with your steps one and two. However, need some more clarification on step 3.
The digits have to be in between 400 and 1000, ie; 400 is excluded. Now, in the step 2, if we consider the digit 0, then we should not consider 0 again in step 3, else we get 400 (if step 1 gives 100th digit as 4). Thus, it should be 3 ways*6 ways*5 ways=90 ways

or an another approach can be to substract just that one case which results into 400. ie; 3ways*6ways*6ways - 1 (the case for 400) = 107

I am not sure of the right answer/approach. I got the question from a textbook according to which the answer is 60. However, that cannot be the case. Would be great, if experts could share their expertise on this.

winniethepooh wrote:With the digits given the numbers that can be made can only be three digits.

The first place can be filled only by 4 or 5 or 6 = 3 ways.
The second place can be filled by any of the 6 digits = 6 ways.
The third place can be filled by any of the 5 remaining digits + the digit used in second place above = 6 ways
Therefore total number of ways = 3 x 6 x 6 =108 ways(numbers)

hundred's digit = 4,5,6 = 3
ten's digit = 6
units digit = 6

thus 3*6*6 = 108

pankajks2010 wrote:
I am not sure of the right answer/approach. I got the question from a textbook according to which the answer is 60. However, that cannot be the case. Would be great, if experts could share their expertise on this.

Hi,
If each digit can be used only once, then the answer will be 60.
a b c
a can be 4,5or6
So, a can be chosen in 3C1 = 3ways
b can be chosen from the remaining 5 digits in 5C1 = 5ways
c can be chosen from the remaining 4 digits in 4C1 = 4ways
So, numbers that can be formed = 3*5*4 = 60

well got surprised till i saw pankaj's reply. I got 60. unless mentioned we cannot use digits repeatedly

Frankenstein wrote:

pankajks2010 wrote:
I am not sure of the right answer/approach. I got the question from a textbook according to which the answer is 60. However, that cannot be the case. Would be great, if experts could share their expertise on this.

Hi,
If each digit can be used only once, then the answer will be 60.
a b c
a can be 4,5or6
So, a can be chosen in 3C1 = 3ways
b can be chosen from the remaining 5 digits in 5C1 = 5ways
c can be chosen from the remaining 4 digits in 4C1 = 4ways
So, numbers that can be formed = 3*5*4 = 60

Frankenstein is right. But the question, as it stands, does not imply that the digits cannot be repeated. 444 is a number within the range that can be made with the digits presented.
As is, the answer should be 107: 3*6*6=108, minus that offending option of 400.

My bad, the answer should be 107!