The arithmetic mean of all the distinct numbers that can be obtained

Question

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, will be

A. 2,222
B. 2,442
C. 2,592
D. 3,333
E. 2,834

gmatophobia · Answer

Number of possible arrangements = 4! /2! = 12

Values = 
1421
1412
1241
1214
1142
1124
4121
4112
4211
2411
2114
2141

Sum = 26664

Average = 26664/12 = 2222

Option A

tgsankar10 · Answer

Available digits: 1 1 2 4

Total number of arrangements possible =\frac{4!}{2!}  (Since arrangements within pair of 1's to be neglected)

No of integers when 4 in thousand place (4xxx):  \frac{3!}{2!}=3

No of integers when 4 in hundredth place (x4xx):  \frac{3!}{2!}=3

No of integers when 4 in tenth place (xx4x):  \frac{3!}{2!}=3

No of integers when 4 in units place (xxx4):  \frac{3!}{2!}=3

This is same for no of ways 2 can be arranged.

ext{Sum of 4 & 2: }

(3*4000+3*400+3*40+3*4)+(3*2000+3*200+3*20+3*2)=3*6000+3*600+3*60+3*6=19998

No of integers when 1 in thousand place (1xxx):  3!=6 . Similarly for all other places its 6 times.

ext{Sum of 1: }6*1000+6*100+6*60+6=6666

Sum=19998+6666=26,664

Average=\frac{26,664}{12}=2222

Answer is A

Sujithz001 · Answer

gmatphobia   Bunuel

Is there any other efficient way to solve this?

Thanks!

aryamanpanda · Answer

while the solution is great, in the interest of time, i think there can be a templatized version.

sum of all distinct numbers formed by re-arranging the numbers can be generalized as :

Sum =   / Number of ways of arrangement

this works for all the values, whether you have repeating or non repeating numbers

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