Let abcd be a general four-digit number and all the digits are non-zer

There are 7 combinations:
abcd
1236
2349
1359
1348
1247
1258
1269
Each can be rearranged in 6 ways. Total 6*7 = 42 .
C.

dnalost I am only getting ways. So total of 36 ways. I guess your one 1349 is not correct one

Could someone help explain this to me, I'm getting 12. The way I thought about it was finding the smallest value for D possible first given a,b,and c have to be unique. This would mean D=1+2+3; D=6.

You can arrange the 1,2,and 3 six different ways (6!) so that's six cases where D=6.

The next smallest value for D is 9 (2+3+4). You can arrange the 2,3,and 4 six ways again so you are at 12 numbers.

1+2+3=6 (1236)
Arranging 1,2,3 so six combinations.

1+2+4=7 (1247) arranging 1,2,4 six combinations

1+2+5 = 8 (1258) again 6 combinations

1+2+6 = 9 (1269) again 6 combinations

1+3+4 =8 (1348) ; 6 combinations

1+3+5 =9 (1359) ; 6 combinations

2+3+4 =9 (2349); 6 combinations

Hence , total 7 × 6 = 42 combinations of such 4 digit nos.


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if d is 3 abc will have to be (111)- >> not possible as number are distinct
if d is 4 abc will have to be 112 or some other combination with 0 -->> Both not possible as neither 0 not repetition is allowed


Like this if we see unique combinations are possible only when d is
6,7,8,9

when d = 6 abc could be (123) -> can be arranged in 3! ways so total 6 ways
when d is 7 abc could be (124) -> can be arranged in 3! ways so total 6 ways
when d is 8 abc could be
(125) -> can be arranged in 3! ways so total 6 ways
(134)-> can be arranged in 3! ways so total 6 ways
when d is 9 abc could be
126 -> arranged in 3! ways = 6 ways
135 -> arranged in 3! ways = 6 ways
234 -> arranged in 3! ways = 6 ways


If we sum all these we get 42 combinations in total

Given abcd is a 4 digit number, such that a,b,c,d are distinct non zero digits & a + b + c = d

Hence lets start from the max value that d can take, d = 9, hence we have {a,b,c} = {6,2,1} or {5,3,1} or {4,3,2}

each combination of {a,b,c} can be arranged among themselves in 3! = 6 ways.

Therefore for d = 9, we have 6 * 3 = 18 ways

Similarly for d =8, we have {a,b,c} = {5,2,1} or {4,3,1}, hence 12 ways

for d = 7, we have {a,b,c} = {4,2,1}, hence 6 ways

& for d = 6, we have {a,b,c} = {3,2,1}, hence 6 ways.

we cannot take d < 6, since we don't have {a,b,c} such that they satisfy a+b+c = d

Hence Total # of 4 digit numbers required = 18 + 12 + 6 + 6 = 42


Answer E.


Thanks,
GyM
_________________

Is there any way to solve it without having to list the possible digits?
For sum =6 and 7, we have 1 combination (and its arrangements), for sum= 8 we have 2 and for sum=9 we have 3.
Is that a rule to predict the number of combinations for each sum, so we can just multiply them by their arrangements of 3!, without having to figure out which they are?
mikemcgarry
Bunuel

Focus on first 3 digits. The last digit will automatically be the sum of first 3.
Since all digits are distinct, the smallest sum of first 3 will be 1 + 2 + 3 = 6.
Since their sum must be a digit, it cannot be more than 9.

The possible combinations with 1 and 2 are:
1, 2, 3
1, 2, 4
1, 2, 5
1, 2, 6 (sum = 9)

The possible combinations with 1 and 3 are:
1, 3, 4
1, 3, 5 (sum = )

There are no possible combinations with 1 and 4 (or higher) since 1, 4, 5 will give 10 as last digit which is not possible.

The possible combinations with 2 and 3 are:
2, 3, 4 (already sum is 9)

Hence there are 7 combinations each of which can be arranged in 3! ways (since each arrangement is a different number)
Total such numbers = 7 * 3! = 42

Answer (E)