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What is the average (mean) of all 5-digit numbers that can be formed [#permalink]
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AkshdeepS wrote:
Archit3110 wrote:
Bunuel wrote:
What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 and exactly once?

(A) 48000
(B) 49999.5
(C) 53332.8
(D) 55555
(E) 56432.8


total terms = 5! = 120
and sum of all terms = fix one place so 4! * ( 1+3+5+7+8)*(11111) = 639996
avg 639996/120 = 53332.8
IMO C


Can you explain this method in detail, Archit?


AkshdeepS

total no of combinations which would be formed using digits 1,3,5,7,8 = 5! = 120
and the series of each integer would be of 120/5 = 24 times or say 4!
also a 5 digit no is of the form 10000+10000+1000+100+10+1= 11111
series of 1,3,5,7,8 can be written as ; 4!*(1+3+5+7+8)*(11111) = 639996
then avg = 639996/120 = ~ 53332.8
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Re: What is the average (mean) of all 5-digit numbers that can be formed [#permalink]
[quote="Archit3110"][quote="AkshdeepS"][quote="Archit3110"][quote="Bunuel"]What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 and exactly once?

(A) 48000
(B) 49999.5
(C) 53332.8
(D) 55555
(E) 56432.8[/quote]

total terms = 5! = 120
and sum of all terms = fix one place so 4! * ( 1+3+5+7+8)*(11111) = 639996
avg 639996/120 = 53332.8
IMO C[/quote]

Can you explain this method in detail, Archit?[/quote]

[url=https://gmatclub.com:443/forum/memberlist.php?mode=viewprofile&un=AkshdeepS][b]AkshdeepS[/b][/url]

total no of combinations which would be formed using digits 1,3,5,7,8 = 5! = 120
and the series of each integer would be of 120/5 = 24 times or say 4!
also a 5 digit no is of the form 10000+10000+1000+100+10+1= 11111
series of 1,3,5,7,8 can be written as ; 4!*(1+3+5+7+8)*(11111) = 639996
then avg = 639996/120 = ~ 53332.8[/quote]

Thanks a lot..Cheers..This is a new learning for me.

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Re: What is the average (mean) of all 5-digit numbers that can be formed [#permalink]
Well,
Since the total terms = 5! = 120
and sum = fix one place so 4! * ( 1+3+5+7+8)*(11111) = 639996
avg 639996/120 = 53332.8
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Re: What is the average (mean) of all 5-digit numbers that can be formed [#permalink]
I understand that there are 120 possible numbers from the 5 digits.

I don't understand anything after that, can someone explain in layman terms?
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What is the average (mean) of all 5-digit numbers that can be formed [#permalink]
2
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Bunuel wrote:
What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, 8 and exactly once?

(A) 48000
(B) 49999.5
(C) 53332.8
(D) 55555
(E) 56432.8


We have to first find out the sum of all the numbers formed by 1,3,5,7 and 8.

Total numbers formed by 1,3,5,7 and 8 = 5! = 120.

Numbers with 8 as unit digits = 4! [ we fix 8 at the end remaining digits can be arranged in 4! ways]

Similarly the no. of numbers with 7,5,3, and 1 as unit digits are 4! for each of them.

So to find the sum of the unit digits of all numbers thus formed we will have to find value of (1+3+5+7+8)*4! =24* 24

Similarly for all others digits places (tens,thousands ,so on) the sum will also be 24*24.

Therefore sum of all the numbers thus formed= ( 10000 * 24*24 + 1000*24*24 + 100*24*24 + 10*24*24 + 24*24)
= 24*24 * 11111
Average = 24 * 24 *11111/120
= 48 * 11111/10
= 53332.8
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What is the average (mean) of all 5-digit numbers that can be formed [#permalink]
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