In how many ways can digits 0, 2, 4, 7 be arranged to make a 4 digit

0 as a digit in this question makes the question interesting.

Answer changes as soon as the "0" is replaced by 2 or any other digit.

Why cant the numbers starting from 0 be considered in this scenario?

I mean 0472 may be a distinct number on a code.

We are NOT looking for 4-digit codes, we are looking for 4-digit numbers, so while a 4-digit code can start with 0 (for example 0123 code is a valid 4-digit code), a 4-digit number cannot start with 0 (0123 is NOT a 4-digit number, it's a 3-digit number 123).

@johneagle
Bunuel is correct, each and every word of a question stem matters. If the word "number" was to be replaced by "codes" or "arrangement of 4 digits", answer would change.

The answer is C: 18
3x3x2x1 = 18

Could you explain the thought process behind this. Why aren't you subtracting the number of 2 digits possible and number of 1 digits possible.

Question : In how many ways can digits 0, 2, 4, 7 be arranged to make a 4 digit number without repetition

We have 4 distinct digits 0,2,4,7 . The different possible 4 digit numbers are

2047,2074,2470,2407,2704,2740
4027,4072,4207,4270,4702,4720
7024,7042,7204,7240,7402,7420 I.e 18 Numbers in all.

An easier way to solve this instead of writing down the numbers would be :

Lets say we have 2 digits say 1,2.
The number of 2 digits numbers possible is 2! ( 12,21)

Lets say we have 3 digits 0,1,2.
The possible number of 3 digits combinations possible is 3! ( 012,021,102,120,210,201 )
But 012 and 021 are not 3 digit numbers right ? They are 2 digit numbers ( 012 is same as 12, 021 is same as 21)
Hence the correct number of 3 digit numbers possible are 3! - 2 or (3! - 2!)

In the original question we have 4 distinct digits 0,2,4,7.
The possible no of 4 digit combination is 4! , but this includes combinations such as 0247,0274 etc,... which we need to subtract from 4!.
To find combinations such as 0274,0247.... we need to start the combination with 0 and the other 3 digits can be placed in any order.
0(2,4,7) - Number of ways to do this is 3!

Hence the num of 4 digits number possible is 4! - 3! = 18.

We don't have to subtract 2 digit or 1 digit number because we are not counting them.
Example : 2 digits number 24 or 47 are not part of our Initial possible set of numbers(4!) :

2047,2074,2470,2407,2704,2740
4027,4072,4207,4270,4702,4720
7024,7042,7204,7240,7402,7420
0247,0274,0427,0472,0724,0742

Hope this helps.

Using FCP:

____ * ____ * ____ * ____

Conditions: 1) must be a 4-digit number; and 2) numbers from which to choose, no repetition: (0, 2, 4, 7)

First slot can be 2, 4 or 7 - there are three possibilities

_3_

Second slot also has only three possibilities; once 2, 4, or 7 is chosen, there are three numbers left

_3_ * _3_

Then, third slot, there are two numbers that remain as possibilities. Fourth slot: one possible number.

_3_ * _3_ * _2_ * _1_

=\(18\)

Answer C

Since the first digit of any number can’t be 0, there are 3 choices for the first digit. Since no digits can be repeated, the number of choices for the second, third, and fourth digits will be 3, 2, and 1, respectively. Thus, the number of ways a 4-digit number can be formed is 3 x 3 x 2 x 1 = 18.

Answer: C

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