Question 535615: Please help me to solve this question :
How many different numbers can be formed by taking one, two, three and four digits from the digits 1, 2, 7 and 8, if repetition are not allowed?
can I do it this way :
4P1 x 4P2 x 4P3 x 4P4
= 4 x 12 x 24 x 24
= 27648
Found 2 solutions by richard1234, htmentor:
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Not quite. You have to add them because you're taking the number of one-digit numbers, plus the number of two digit numbers, etc. Hence it would be 4+12+24+24 = 64.
Answer by htmentor(1343)  (Show Source):
You can put this solution on YOUR website! You are correct in assuming this is a permutation problem. However, to find the
total number, the operation required is addition, not multiplication:
N = P(4,1) + P(4,2) + P(4,3) + P(4,4) = 4 + 12 + 24 + 24 = 64
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