What is sum of all rearrangements of the 4-digit number 3321?

Question

A. 27,777
B. 28,888
C. 29,997
D. 29,999
E. 39,999

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nick1816 · Accepted Answer

Total possible arrangements  =\frac{ 4!}{2!} = 12

Probability of 3 appears at any position is two times as compared to that of 1 and 2.

Hence 3 will appear 6 times and 1 and 2 will appear 3 times at each position.

sum of all rearrangements  = 1111*(6*3+3*2+3*1) = 1111*27 = 29997

nick1816 · Answer

Sum of digits at unit place or at tens place or at hundreds place or at thousands place of all rearrangements = (6*3+3*2+3*1) =27

Sum of all rearrangements = (1000*27)+(100*27)+(10*27)+27 = 27*(1000+100+10+1) = 27*1111

poojakhanduja3017 · Answer

Can you explain how you came at this formula for sum of arrangements?

preetamsaha · Answer

xyzw 
case I : w=1, (x,y,z)= (3,3,2) ; no of cases after all arrangements= 3!/2! =3
case II : w=2, (x,y,z)= (3,3,1) ; no of cases after all arrangements= 3!/2! =3
case III : w=3, (x,y,z)= (1,3,2) ; no of cases after all arrangements= 3! =6

this is the same scenario for all x,y,z.

Therefore, sum of units place = 3*1 +3*2 + 6*3 = 27.
Sum of tens place = 270
Sum of hundreds place = 2700
Sum of thousands place =27000

Sum of all arrangements of 3321 = 29997.
 
correct answer C

ScottTargetTestPrep · Answer

The number of ways to arrange the digits 1, 2, 3, and 3 is 4!/2! = 12. Of these 12 numbers, digits 1 and 2 each appear as the units digit 3 times, and digit 3 appears 6 times (because there are two 3’s); therefore, the sum of all the units digits is (1 + 2) x 3 + 3 x 6 = 9 + 18 = 27. Using the same analogy, these digits should yield a sum of 270 when they are the tens digit (since they will appear the same number of times as when they are the units digit, only ten times as much as the sum), 2700 when they are the hundreds digit and 27,000 when they are the thousands digit. Therefore, the sum of the 12 numbers is:

27 + 270 + 2700 + 27,000 = 29,997

Answer: C

sambitspm · Answer

I thought of extending the formula -  (a+b+c+d+....)* (1111....ntimes)* ((n-1)!)  
 
Here the case is 3321
So by formula = (3+3+2+1) * (1111) * 3! should be sum

WAIT

3 is repeated 2 times

so Sum =   / 2!
 = 9 * 1111 * 6 / 2 = 27*1111 = 29997

C

I confirmed the theory with 33311
Cases:
33311
33131
33113
31331
31133
31313
13331
13313
13133
11333

Total comes as 244442

BY formula = (3+3+3+1+1) * 11111 * 4! / (3! * 2!)
= 244442

Kinshook · Answer

What is sum of all rearrangements of the 4-digit number 3321?

Total number of rearrangements = 4!/2! = 12

3321
3231
2331
3312
3132
1332
1233
1323
2133
2313
3123
3213

Sum = 29997

IMO C

Adit_ · Answer

Idk if this is a fruitful approach but what I did was:

Calculate units digit value and then take it from there:

for starting with 1st digit as 3 we have: 1+2+3 = 6*2= 12( this is because each unit digit here repeats twice)

For remaining:
3+3+1=7
2+3+3=8
Therefore:
12+7+8=27 and we know its a value ending with 7

Only option A and C is valid now

We now see 3000*6+1000*3+2000*3=27000 (very rough estimate of total value taking only 1st digit)

When the 1st digit was 3: 300 repeats twice 200 repeats twice (rough rough estimate again) 
Here we already see value exceeding 27777 which is A

Therefore by POE it has to be C.

What is sum of all rearrangements of the 4-digit number 3321?

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What is sum of all rearrangements of the 4-digit number 3321?

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