So I'll try to type and figure out this problem at the same time. It has to be less than 10,000, so it's going to be a number with one, two, three, or four digits.
* One Digits * 5 (1 Total)
* Two Digits * 50, 41, 32, 23, 14 (6 total)
* Three Digits * 500, 410, 401, 320, 302, 311, 230, 203, 221, 212, 140, 104, 113, 131, 122 (21 Total)
* Four Digits * 5000, 4100, 4010, 4001, 3200, 3020, 3002, 3110, 3101, 3011, 2300, 2030, 2003, 2120, 2102, 2210, 2201, 2120, 2102, 2111, 1400, 1040, 1004, 1130, 1103, 1220, 1202, 1310, 1301. (51 Total)
B. 51 (wrong), there should be a short cut to a correct answer?
Combinations
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vvsmart
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Only 1 non-zero digit - 4
5, 50,500,5000 - 4
two non-zero digits - 24
23 - 2 ( 23 & 32)
230 - 4 (230, 203, 320, 302)
2300 - 6 (2003, 2030, 2300, 3200, 3020, 3002)
Total = 12
or calculate overall as 4!/2! = 12 ( find combinations for 4 digits where 2 are repeating 2, 3, 0, 0)
Similarly 12 options for 1,4,0,0
3 non-zero digits - 24
1,2,2,0
Total = 4!/2! = 12
1,3,1,0
Total = 4!/2! = 12
4 non-zero digits = 4
1,1,1,2
Total = 4!/3! = 4
Overall total = 56
5, 50,500,5000 - 4
two non-zero digits - 24
23 - 2 ( 23 & 32)
230 - 4 (230, 203, 320, 302)
2300 - 6 (2003, 2030, 2300, 3200, 3020, 3002)
Total = 12
or calculate overall as 4!/2! = 12 ( find combinations for 4 digits where 2 are repeating 2, 3, 0, 0)
Similarly 12 options for 1,4,0,0
3 non-zero digits - 24
1,2,2,0
Total = 4!/2! = 12
1,3,1,0
Total = 4!/2! = 12
4 non-zero digits = 4
1,1,1,2
Total = 4!/3! = 4
Overall total = 56
