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primary 5 | Maths
| Decimals
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Date Posted: 4 years ago
Views: 844
For your level, you should list down every single possibility.
These are the possible 3-digit combinations from digits 2, 4, 7 and 8 without repetition of digits.
Digits starting with 2 are 247, 248, 274, 278, 284 and 287.
Digits starting with 4 are 427, 428, 472, 478, 482 and 487.
Digits starting with 7 are 724, 728, 742, 748, 782 and 784.
Digits starting with 8 are 824, 827, 842, 847, 872 and 874.
There are a total of 24 possible combinations, two of which are 247 and 274. Therefore, the remaining 22 combinations are possible.
These are the possible 3-digit combinations from digits 2, 4, 7 and 8 without repetition of digits.
Digits starting with 2 are 247, 248, 274, 278, 284 and 287.
Digits starting with 4 are 427, 428, 472, 478, 482 and 487.
Digits starting with 7 are 724, 728, 742, 748, 782 and 784.
Digits starting with 8 are 824, 827, 842, 847, 872 and 874.
There are a total of 24 possible combinations, two of which are 247 and 274. Therefore, the remaining 22 combinations are possible.
See 2 Answers
For the first digit, he can choose one digit out from 2, 4,7 or 8. Hence he has 4 choices.
For the second digit, he was left with 3 choices as one digit has been selected and there are no repeatation of digits.
For the last digit, he was left with 2 choices as two digits had been selected.
Total combinations of digits =4*3*2=24
Since, he already tried 2 combination,
Maximum more number of combinations that he can try=24-2=22
For the second digit, he was left with 3 choices as one digit has been selected and there are no repeatation of digits.
For the last digit, he was left with 2 choices as two digits had been selected.
Total combinations of digits =4*3*2=24
Since, he already tried 2 combination,
Maximum more number of combinations that he can try=24-2=22
done
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Date Posted:
4 years ago
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