Question on Permutation and Combination

Topic : Word Problem on arragement of five digit numbers

Question : Samer formed all possible five-digit numbers with no two digits are repeated, using members of the set{1,2,3,4,5,6,7}.These numbers are listed in the ascending order and this list of numbers is divided in the middle into two parts. What are the middle three digits of the last number in the first part?

Answer :

We have to find all possible 5 digit numbers, using the given 7 digits. There are 7P5 = 7!/ 2! = 2520 possible numbers.
If we divide these into two equal parts, there are 1260 numbers in each part. So we have to find the last number of the first part: which is the 1260th number.
Out of the 2520 numbers, the numbers starting with each of the digits (1,2 ,….7) are equal = 2520/7 = 360. That is, there are 360 numbers starting with 1, another 360 starting with the digit 2 and so on.
Thus as wew arranging the numbers in the ascending order, the first 360 numbers start with 1. The numbers 361st to 720th start with the digit 2.
721st to 1080th start with the digit 3.
1081st to 1440th number start with 4. Therefore the 1260th number starts with the digit 4.
The second digit may be any of the six digits 1,2,3,5,6 or 7. There are 360/6 = 60 numbers with any of these digits as the second digit in the five digit number. So 1081st to 1140th (that is 1080 +60) have the second digit as 1.
1141st to 1200th number have the second digit 2
1201st to 1260th number have the second digit 3.
We can understand that 1260th number starts with 43 and we also know that this is
the largest number with 43 as the first two digits. The other digits that we can use to make the five digit number are 1,2,5,6,7. So we can see that the largest number possible is 43765.

We want the three digits at the middle of the number 43765, which is 376.
Answer is 376

Another way to Solve the Problem.

1-6*5*4*3 = 360
2---------=360
3---------=360
4---------=360
5---------=360
6---------=360
7---------=360



the first digit has to be 4 since all numbers starting with 4 are in the middle

four positions left numbers are 1,2,3,5,6,7

- - - -

1 6*5*4= 120
2........120
3........120
4 not possible as it has occupied first position already and digits cant repeat
5........120
6........120
7........120

from this second position has to be either 3 or 5
first is 4 second is either 3 or 5


4 3 5*4*3= 60
4 5 5*4*3= 60

total are 360*7 = 2520

the firsr split is 1260 the number is what we need
360+360+360= 1080
so the next 180 should be in the first half of the numbers starting with 4
so 4 3 is the right combination

since it is the last number it should be highest possible number

which is

4 3 7 6 5
so answer is the 376 are the middle two digits.

Hope any of the method will be helpful for you to solve similar kind of Questions.