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19 Oct 2022, 06:49
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Here's my method of solving without formula.
So, there are 24 integers i.e., every digit will have 6 possibilities. How?
1234
1243
1324
1342
1423
1432
Similar for 2 , 3 ,and 4
So the sum of 1 will be = 1000+1000+1000+1000+1000+1000 = 6000 ish ( actually greater than 6000)
Similarly for 2 will be = 2000*6 = 12000 ish ( actually greater than 12000)
For 3 will be = 3000*6 = 18000 ish ( actually greater than 18000)
For 4 will be = 4000*6 = 24000 ish ( actually greater than 24000)
So overall the sum of 24 integers will be 6000+12000+18000+24000= 60000 ( greater than 60,000)
Hence Choice E
09 Mar 2023, 23:20
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To find the sum of all 24 integers formed by using each of the digits 1, 2, 3, and 4 exactly once, we can use a bit of logic. Since each digit will appear in each position (units, tens, hundreds, and thousands) equally, we can start by finding the sum of all the possible permutations of the digits.
To do this, we can use the formula:
n!
where n is the total number of items (in this case, 4), and r is the number of items we are selecting at a time (also 4, since we are using all four digits in each number).
So the total number of possible permutations of the digits is:
4! = 24
Each of these permutations can be used as the units digit in one of the 24 numbers in the sum. So to find the sum of all 24 numbers, we simply need to add up the permutations in each of the other three positions (tens, hundreds, and thousands).
Since each digit appears in each position equally, the sum of all the permutations in each position will be the same. So we can find the sum of all the permutations in each position by adding up the digits (1+2+3+4=10) and multiplying by the number of permutations:
10 * 6 = 60
So the sum of all 24 integers formed by using each of the digits 1, 2, 3, and 4 exactly once is:
60 + 600 + 6000 + 60000 = 66660
To do this, we can use the formula:
n!
where n is the total number of items (in this case, 4), and r is the number of items we are selecting at a time (also 4, since we are using all four digits in each number).
So the total number of possible permutations of the digits is:
4! = 24
Each of these permutations can be used as the units digit in one of the 24 numbers in the sum. So to find the sum of all 24 numbers, we simply need to add up the permutations in each of the other three positions (tens, hundreds, and thousands).
Since each digit appears in each position equally, the sum of all the permutations in each position will be the same. So we can find the sum of all the permutations in each position by adding up the digits (1+2+3+4=10) and multiplying by the number of permutations:
10 * 6 = 60
So the sum of all 24 integers formed by using each of the digits 1, 2, 3, and 4 exactly once is:
60 + 600 + 6000 + 60000 = 66660
vivek1810971
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11 Mar 2023, 01:33
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This is how i solved it:
I saw the answer options first. Observed that there is sufficient difference between each answer choice.
Next i proceeded to find an approximate answer. Following calculation shows the method i used:
As it is given that it is a four digit integer made out of the following digits: 1,2,3 and 4.
_ _ _ _ : fill the left most blank with 1 first.
1 _ _ _ : in how many different ways can the remaining three digits be chosen, in 3! factorial ways. Hence there are 6 numbers which have 1 in thousands place.
Therefore are 6 numbers that are atleast above 1000. So counting 1000 x 6 : 6000. (the actual sum of these numbers will be more than 6000)
Applying the same logic by picking 2,3 & 4 as the thousands digit:
2 _ _ _: 3! factorial ways to fill the rest of the digits. Hence atleast 6 numbers greater than 2000. Hence, 2000 x 6 = 12000.
3 _ _ _: 3000 x 6 = 18000.
4 _ _ _: 4000 x 6 = 24000.
Total sum : 6000 + 12000 + 18000 + 24000 = 60,000.
We also know that the actual sum will be greater than 60,000 and there is only one option that is greater than 60,000. Hence, option E.
I hope the explanation helps.
I saw the answer options first. Observed that there is sufficient difference between each answer choice.
Next i proceeded to find an approximate answer. Following calculation shows the method i used:
As it is given that it is a four digit integer made out of the following digits: 1,2,3 and 4.
_ _ _ _ : fill the left most blank with 1 first.
1 _ _ _ : in how many different ways can the remaining three digits be chosen, in 3! factorial ways. Hence there are 6 numbers which have 1 in thousands place.
Therefore are 6 numbers that are atleast above 1000. So counting 1000 x 6 : 6000. (the actual sum of these numbers will be more than 6000)
Applying the same logic by picking 2,3 & 4 as the thousands digit:
2 _ _ _: 3! factorial ways to fill the rest of the digits. Hence atleast 6 numbers greater than 2000. Hence, 2000 x 6 = 12000.
3 _ _ _: 3000 x 6 = 18000.
4 _ _ _: 4000 x 6 = 24000.
Total sum : 6000 + 12000 + 18000 + 24000 = 60,000.
We also know that the actual sum will be greater than 60,000 and there is only one option that is greater than 60,000. Hence, option E.
I hope the explanation helps.
