Combinatorics 02

_ _ _ _

The first can be filled in 7 ways (excluding the other 3 Nos)

Second can be filled in 8 ways (Excluding 3rd & 4th but including 0)

For the No. to be divisible by 4, last 2 Nos shld be divisible by 4, so you have 96/4 = 24 possibilities.

But 2 Nos cannot be the same so we exclude 44 and 88. So 22 options remain for the last 2 Nos.

Multiplying we have 7*8*22 = 1232

THis also includes cases in which you have 2 0's. (i.e the 2nd is 0 and either of 3rd or 4th is 0)

To exclude those cases; consider the second position takes 0

S0,

_ 0 _ _

Now the first can take 7 values as earlier.

second takes 0

The third and fourth places cannot take 0 but should be divisible by 4 (exclude values 04,08,60,20,80,40)

SO thats 22-6 = 16

So we have 7 * 1 * 16 = 112 cases where the second place takes 0.

these are duplicates and hence we have 1232-112 = 1120

IMO [spooiler]C[/spoiler]

Too long to be a GMAT sum though.

i think answer shd be 7x8x22
becz only 22 pairs are possible that r div by 4

I would split this into two cases:
1. The last 2 digits contain a zero (04, 08, 20, 40, 60, 80) --> 6 choices
2. The last 2 digits do not contain a zero (12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96) --> 16 choices

Case 1:
6 choices for last 2, 8 choices for thousands digit, 7 choices for hundreds digit:
6*8*7

Case 2:
16 choices for last 2, 7 choices for thousands digit, 7 choices for hundreds digit:
16*7*7

Answer: 6*8*7 + 16*7*7
This is pretty close to 22*8*7 but it's a bit lower, because you shouldn't be counting things like 0532 as answers.

The main lesson from all of these questions that keep getting posted is that you need to think about the situations where zero is a possible choice for the first digit differently than the cases where zero is already used and therefore is NOT a possible choice for the first digit (in this case the thousands digit).

knight247 wrote:How many even four digit numbers can be formed by using the digits 0-9 which are divisible by 4 and no two digits are repeated?
A 336
B 784
C 1120
D 1804
E 1936

Don't have an OA. Detailed explanations would be appreciated

Good = Total - Bad.

Total options if we disregard whether 0 appears in the thousands place:
The 2 rightmost digits must form a multiple of 4.
The total number of 2-digit multiples of 4 = 25.
Of these 25 multiples of 4, 3 options -- 00, 44 and 88 -- are not allowed.
Thus:
Number of options for the 2 rightmost digits = 25-3 = 22.
Number of options for the thousands digit = 8. (Any digit 0-9 other than the 2 digits already used.)
Number of options for the hundreds digit = 7. (Any digit 0-9 other than the 3 digits already used.)
Total options = 22*8*7 = 1232.

Bad options:
A bad option puts 0 in the thousands place.
We need to count how many ways the remaining 3 digits can be chosen.
Since 0 is in the thousands place, the last 2 digits cannot include 0.
Thus, of the 22 multiples of 4 considered above for the 2 rightmost positions, 6 of these options -- 04, 08, 20, 40, 60, and 80 -- are not allowed here.
Thus:
Number of options for the 2 rightmost digits = 22-6 = 16.
Number of options for the hundreds digit = 7. (Any digit 1-9 other than the 2 digits used in the 2 rightmost positions).
Bad options = 16*7 = 112.

Good = 1232-112 = 1120.

The correct answer is C.