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Question 1024301: Hi my beloved tutors, I need help on solving this problem,
could you help me? Thankss
Consider the set of all the 7-digit numbers that can be
obtained using, for each number, all the digits 1, 2, 3,...,7.
List the numbers of the set in increasing order and split the
list exactly at the middle into two parts of the same size.
What is the last number of the first half?
(A) 1234567 (B) 3765421 (C) 4123567 (D) 4352617 (E) 4376521
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website!
There are 7! = 5040 numbers in the list.
We want to know what the 5040/2 = 2520th
number is. So we seek arrangement #2520
in the list.
7! = 7*6!. There are 7 groups of 6! beginning
with each of the 7 digits. 6! = 720. Therefore,
the 1st 6! begin with 1. That's #1 through #720.
the 2nd 6! begin with 2. That's #721 through #1440.
the 3rd 6! begin with 3. That's #1440 through #2160.
the 4th 6! begin with 4. That's #2161 through #2880.
So #2520 will be in the middle of the 4th 6!
We examine the 6! that begin with 4.
6! = 6*5!. There are 6 groups of 5! beginning
with 4 which have as 2nd digit each of the 6 digits
1,2,3,5,6,7. 5! = 120. Therefore
the 1st 5! begin with 41. That's #2161 through #2280.
the 2nd 5! begin with 42. That's #2281 through #2400.
the 3rd 5! begin with 43. That's #2401 through #2520.
Aha! #2520 happens to be the very one we are looking
for. And it is the last and largest one on the list
which begins with 43. That would have to be 4376521
Answer: 4376521, choice E.
Edwin
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