How many 4 digit numbers are there, if it is known that the first digi

Question

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

A. 20
B. 150
C. 225
D. 300
E. 320

A. 20
B. 150
C. 225
D. 300
E. 320

Bunuel · Accepted Answer

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?  
A. 20
B. 150
C. 225
D. 300
E. 320

4 options for the first digit: 2, 4, 6, 8;
5 options for the second digit: 1, 3, 5, 7, 9;
4 options for the third digit: 2, 3, 5, 7;
4 options for the fourth digit: 0, 3, 6, 9.

Four digit # possible without the restriction (about the digit 2): 4*5*4*4=320

Numbers with two 2-s, 2X2X 1*5*1*4=20.

Thus there are 320-20=300 such numbers.

Answer: D.

Archit3110 · Answer

target find how many 4 digit numbers are there such that 1st digit is even , 2nd odd , 3rd prime & 4th is divisible by 3
1st digit (0,2,4,6,8)
2nd digit ( 1,3,5,7,9)
3rd digit ( 2,3,5,7)
4th digit ( 0,3,6,9)
condition that 2 can be used once in same number 
case 1: first digit is 2 
1*5*3*4 ; 60
case 2 : first digit is not 2
3*5*4*4 ; 240
total cases : 240+60 ; 300
option D

rgreenhill · Answer

Possible numbers in each column

1: 2,4,6,8 (cannot be 0 because has to be 4 digit number)
2. 1,3,5,7,9
3. 2,3,5,7
4. 0,3,6,9 (it's easy to forget the 0 here)

It is now easiest to imagine 3 scenarios:

1. a 2 appears in the first column and no other column. 
That means there is only 1 way of arranging the first column, 5 ways the second, 3 ways the third, and 4 ways the fourth.
1x5x3x4 =60

2. a 2 appears in the third column and no other column
That means there are 3 ways of arranging the first column, 5 ways the second, 1 way the third, and 4 ways the fourth.
3x5x1x4 =60

3. no 2s appear at all
That means there are 3 ways of arranging the first column, 5 ways the second, 3 ways the third, and 4 ways the fourth.
3x5x3x4 =180

Next, add all the scenarios:
60+60+180 =300

D

SahilGera18 · Answer

As per me the answer should be 225 as in the fist digit 0 cannot come .

So i have done following:-

1st Digit--> 2,4,6,8
2nd Digit-->1,3,5,7,9
3rd Digit-->2,3,5,7
4th Digit--> 3,6,9

So if i fix 2 in 1000th positon i have 5X3X3=45 possibility.
now if i Fix 4 i have--> 5x4x3=60
Same for 6&8-60

total-45+60+60+60=225

rgreenhill · Answer

I think you forgot to include the 0 in the units column, because there are always 4 numbers there that are divisible by 3: 0,3,6,9

Do you mind sharing why you fixed the 4, 6, and 8?

SahilGera18 · Answer

I fixed 4,6,8 because the thousand have to even so total 2,4,6,8

Posted from my mobile device

sujoykrdatta · Answer

Possibilities for the first digit = 2, 4, 6, 8
Possibilities for the second digit = 1, 3, 5, 7, 9
Possibilities for the third digit = 2, 3, 5, 7
Possibilities for the fourth digit = 0, 3, 6, 9

Total possibilities = 4 * 5 * 4 * 4 = 320
Number of cases where both the 1st digit and the 3rd digit is 2 = 1 * 5 * 1 * 4 = 20

Thus, number of cases where 2 is used only once = 320 - 20 = 300
Answer D

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How many 4 digit numbers are there, if it is known that the first digi

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