- #1
Sleek
- 60
- 0
Hello,
Recently, I've come across a textbook question based on permutation. I've even got the answer to that, but i want to verify my steps, because each time i solved the question, i realized how tricky it was, and i had to modify my approach. Heres my final approach, please spot any mistakes you find.
Question : There are 9 digits, namely {1,2,3,4,5,6,7,8,9}. The thing that has to be found out is, the number of four digit numbers which can be constructed from those digits such that no digits are repeated in the number and all the numbers are above/greater than 3400. This sum is quite tricky, as if it would have been a number like 4000, then it might be easy to use the Fundamental Principle.
Answer: : Heres my attempted answer.
By using the digits from 1 to 9 and between 3000 and 4000, the least number above 3400, without any repitition of digits which can be constructed is 3412. The greatest is 3987.
From the above two numbers, we can see that,
-> The first place (from the left) only assumes the digit 3. So the number of digits it can be substituted with is 1.
-> The second place, can assume 4,5,6,7,8,9. So the number of digits it can be substituted with is 6.
-> The third place assumes digits from 1 to 8. So no. of combinitions is 8.
-> The fourth place assumes digits 2,3,4,5,6,7. So no. of combinitions is 6.
By using fundamental principle, all four digits numbers between 3000 and 4000 such that, they only have digits 1 to 9, no digit is repeated, is 1*6*8*6 = 288.
Now let's consider all digits above 4000. The reason this is done seperately is because, between 3000 and 4000, the second or/and other places can assume less no. of digits as opposed to cases above 4000.
For all no. above 4000,
-> The first place (from left) can only assume digits 4,5,6,7,8,9. So no. of combinitions is 6.
-> The second, third and fourth place can assume digits 8,7,6 as repitition of digits is not allowed.
So by fundamental principle, no of possible digits above 4000, with no digits repeated is 6*8*7*6 = 2016.
If we add up 288 and 2016, we get 2340, which seems to be the required answer for all possible permutations of 4 digit number using the digits 1 to 9, with no repitition of digits allowed in the number, and above 3400.
The Above 4000 Part was a sum by itself, and the answer to that part is right. But i am concerned about my approach of seperating the sections in 3000 to 4000 and above 4000. Can anyone verify my method?
Regards,
Sleek.
Recently, I've come across a textbook question based on permutation. I've even got the answer to that, but i want to verify my steps, because each time i solved the question, i realized how tricky it was, and i had to modify my approach. Heres my final approach, please spot any mistakes you find.
Question : There are 9 digits, namely {1,2,3,4,5,6,7,8,9}. The thing that has to be found out is, the number of four digit numbers which can be constructed from those digits such that no digits are repeated in the number and all the numbers are above/greater than 3400. This sum is quite tricky, as if it would have been a number like 4000, then it might be easy to use the Fundamental Principle.
Answer: : Heres my attempted answer.
By using the digits from 1 to 9 and between 3000 and 4000, the least number above 3400, without any repitition of digits which can be constructed is 3412. The greatest is 3987.
From the above two numbers, we can see that,
-> The first place (from the left) only assumes the digit 3. So the number of digits it can be substituted with is 1.
-> The second place, can assume 4,5,6,7,8,9. So the number of digits it can be substituted with is 6.
-> The third place assumes digits from 1 to 8. So no. of combinitions is 8.
-> The fourth place assumes digits 2,3,4,5,6,7. So no. of combinitions is 6.
By using fundamental principle, all four digits numbers between 3000 and 4000 such that, they only have digits 1 to 9, no digit is repeated, is 1*6*8*6 = 288.
Now let's consider all digits above 4000. The reason this is done seperately is because, between 3000 and 4000, the second or/and other places can assume less no. of digits as opposed to cases above 4000.
For all no. above 4000,
-> The first place (from left) can only assume digits 4,5,6,7,8,9. So no. of combinitions is 6.
-> The second, third and fourth place can assume digits 8,7,6 as repitition of digits is not allowed.
So by fundamental principle, no of possible digits above 4000, with no digits repeated is 6*8*7*6 = 2016.
If we add up 288 and 2016, we get 2340, which seems to be the required answer for all possible permutations of 4 digit number using the digits 1 to 9, with no repitition of digits allowed in the number, and above 3400.
The Above 4000 Part was a sum by itself, and the answer to that part is right. But i am concerned about my approach of seperating the sections in 3000 to 4000 and above 4000. Can anyone verify my method?
Regards,
Sleek.
.
