For each of the four subsets of the two properties (a) and (b), count the number of four-digit numbers whose digits are either 1,2,3,4, or 5: (a) the digits are distinct. (b) the number is even.
Answers
This question is related to r-permutations from n distinct objects. The only pitfall often overlooked is that those 4-digit numbers with a leading ‘0’ must be excluded because those numbers are not valid 4-digit numbers. From elementary combinatorics course we know the number of such r-permutations is P(n,r)=n!(n−r)!=n×(n−1)×⋯×(n−r+1)P(n,r)=n!(n−r)!=n×(n−1)×⋯×(n−r+1).
Therefore, the number of all 4-digits numbers with distinct digits are P(10,4)=10!(10−4)!=10×9×8×7=5040P(10,4)=10!(10−4)!=10×9×8×7=5040. Now let us exclude those invalid 4-digit numbers that start with ‘0’. This amounts to fixing a ‘0’ in the leftmost position and fill up the remaining 3 positions with 9 distinct digits, which leads to P(9,3)=9×8×7=504P(9,3)=9×8×7=504 numbers.
Subtracting 504504 from 50405040 gives the final answer : 5040−504=45365040−504=4536.
Answer:
Step-by-step explanation:
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