A four-digit number is formed using the digits 3,4,6 and 7 without rep

Question

A four-digit number is formed using the digits 3,4,6 and 7 without repetition. All the possible numbers so formed are added and the sum is equal to P. Which of the following is the value of P?

A. 133200
B. 123320
C. 133220
D. 133320
E. None of these

KarishmaB · Accepted Answer

Using digits 3, 4, 6 and 7, you can make 4! = 24 numbers without repetition. 
Since the probability of each digit in any place is the same as the probability of any other digit in that place, we will have 6 numbers with 3 in units place, 6 numbers with 4 in units place, 6 numbers with 6 in units place and 6 numbers with 7 in units place. 
Same for tens place, hundreds place and thousands place.

4673
4763
...
6734
7634
...
...

Sum = 3*6 + 4*6 + 6*6 + 7*6 + 10*(3*6 + 4*6 + 6*6 + 7*6) + 100*(3*6 + 4*6 + 6*6 + 7*6) + 1000*(3*6 + 4*6 + 6*6 + 7*6) 
Sum = 6*20 *(1 + 10 + 100 + 1000) = 120 * 1111 
Sum = 133,320

Answer (D)

chetan2u · Answer

Now, total numbers that can be formed =  4!=24 .
As, all 4 digits are different, each digit will come  \frac{24}{4}=6  times at each place, that is  at ones, tens, hundreds and thousands.

So  SUM = (1000+100+10+1)*6*(3+4+6+7)=1111*6*20=133320

D

sahibr · Answer

We have four digits ie. 3, 4, 6 and 7.

The four digits can be arranged(without repetition) in 4! ways ie. 24 ways.

Now, just to make the solution more comprehensive, let's write down a few of these numbers.

First lets fix the first digit as 3. Possible numbers are:

3467
3476
3647
3674
3746
3764

We can see that each digit repeats twice in unit's place, twice in ten's place and twice in hundred's place.
When we change the first digit from 3 to 4, 6 and 7, a similar pattern is observed.

Hence when 3 is fixed 4, 6 and 7 repeat twice at each location.
When 4 is fixed, 3,6 and 7 repeat twice.
When 6 is fixed, 3,4 and 7 repeat twice
When 7 is fixed, 3,4 and 6 repeat twice.

All in all, all digits are repeated six times.

Hence 3x6+4x6+6x6+7x6=120 --->(1)

Imagine, all possible combinations lined up vertically to be added.

Starting at unit's place, we leave 0 and carry over 12 to tens place (from (1) we know we get 120)

At tens place, since the same digits have been repeated, we get 120 + 12 from carryover. Hence 132.
We leave 2 and carry over 13.

At hundreds place, again we get 120 + 13. Hence 133.
Leave 3 and carry over 13.

At thousands place, we get 120 + 13. Hence 133.
Since we dont need to carry over any further, we have arrived at our number.

Ans: 133320

Abi1995 · Answer

I used this to solve

Sum = 3*6 + 4*6 + 6*6 + 7*6 + 10*(3*6 + 4*6 + 6*6 + 7*6) + 100*(3*6 + 4*6 + 6*6 + 7*6) + 1000*(3*6 + 4*6 + 6*6 + 7*6) 
Sum = 6*20 *(1 + 10 + 100 + 1000) = 120 * 1111 
Sum = 133,320

Posted from my mobile device

bumpbot · Answer

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A four-digit number is formed using the digits 3,4,6 and 7 without rep

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