The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, 6 taken all at a time is
A. 432
B. 108
C. 36
D. 18
Answer
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Hint: Let us go through by using permutations to find total numbers of digits formed and then calculating the sum of unit place digits.
Complete step-by-step answer:
Here we will first see if any repetitions of digits are allowed or not. According to the question, all the digits have to be taken at a time, so we see that repetition is not allowed. Now, to find the total number of numbers formed by the digits, we will fix one of the digits and calculate all the numbers formed for that digit.
Let us fix 4 at unit place and we will calculate all the numbers that are formed for 4 at unit place.
To calculate the total numbers, we will use the formula \[{}^n{P_r}\], where \[{}^n{P_r}\] = \[\dfrac{{{\text{n!}}}}{{{\text{(n - r)!}}}}\]
where n is the number of digits and r is the number of places to be filled. As, the unit place is filled with 4, so only three places are left and 3 digits are left. Three places (Tens, Hundreds and Thousands) are to be filled with 3, 5, 6. So, we will use the formula of permutation. Here n = 3 and r = 3.
So, total numbers formed with 4 at unit place = \[{}^n{P_r}\] = \[{}^3{P_3}\] = \[\dfrac{{{\text{3!}}}}{{{\text{(3 - 3)!}}}}\] = \[\dfrac{{{\text{3!}}}}{{{\text{0!}}}}\] = 6
Similarly, with 3, 5 and 6 at unit place 6 numbers are formed in each case.
Now, to get a sum of digits at unit place we have to calculate the sum by placing the digit at unit place and multiplying it by numbers formed.
Therefore, sum by placing 4 at unit place = 4 x 6
Similarly sum by placing 3, 5, 6 at unit place will be 3 x 6, 5 x 6, 6 x 6 respectively. So, we will add all the sums to get the required sum.
Sum of digits at unit place of all numbers = 3 x 6 + 4 x 6 + 5 x 6 + 6 x 6 = 6(3+4+5+6) = 6(18) = 108
So, required sum = 108. Option (B) is correct.
Note: For this type of problem it is recommended to use the formula always. There is another method to solve the problem which is by writing all the combinations made by placing numbers at different positions. But this method is very tedious and complicated and takes more time from the method which uses formula. Also, many mistakes are done using this formula.
Complete step-by-step answer:
Here we will first see if any repetitions of digits are allowed or not. According to the question, all the digits have to be taken at a time, so we see that repetition is not allowed. Now, to find the total number of numbers formed by the digits, we will fix one of the digits and calculate all the numbers formed for that digit.
Let us fix 4 at unit place and we will calculate all the numbers that are formed for 4 at unit place.
To calculate the total numbers, we will use the formula \[{}^n{P_r}\], where \[{}^n{P_r}\] = \[\dfrac{{{\text{n!}}}}{{{\text{(n - r)!}}}}\]
where n is the number of digits and r is the number of places to be filled. As, the unit place is filled with 4, so only three places are left and 3 digits are left. Three places (Tens, Hundreds and Thousands) are to be filled with 3, 5, 6. So, we will use the formula of permutation. Here n = 3 and r = 3.
So, total numbers formed with 4 at unit place = \[{}^n{P_r}\] = \[{}^3{P_3}\] = \[\dfrac{{{\text{3!}}}}{{{\text{(3 - 3)!}}}}\] = \[\dfrac{{{\text{3!}}}}{{{\text{0!}}}}\] = 6
Similarly, with 3, 5 and 6 at unit place 6 numbers are formed in each case.
Now, to get a sum of digits at unit place we have to calculate the sum by placing the digit at unit place and multiplying it by numbers formed.
Therefore, sum by placing 4 at unit place = 4 x 6
Similarly sum by placing 3, 5, 6 at unit place will be 3 x 6, 5 x 6, 6 x 6 respectively. So, we will add all the sums to get the required sum.
Sum of digits at unit place of all numbers = 3 x 6 + 4 x 6 + 5 x 6 + 6 x 6 = 6(3+4+5+6) = 6(18) = 108
So, required sum = 108. Option (B) is correct.
Note: For this type of problem it is recommended to use the formula always. There is another method to solve the problem which is by writing all the combinations made by placing numbers at different positions. But this method is very tedious and complicated and takes more time from the method which uses formula. Also, many mistakes are done using this formula.
Last updated date: 20th Sep 2023
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