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28 Mar 2011, 01:54
Kudos
Bookmarks
using the digits 2,3,4,5 and 6 find the sum of all 5 digit numbers that can be formed such that no two digits are the same
The solution given is (2+3+4+5+6)*(11111)*4!
Why is it 4! ??
I think the solution is (2+3+4+5+6)*(11111)* 5!
The number of ways in which 5 digits can be arranged into 5 digit numbers without repetition is 5!.
Each digit will repeat in each of the positions once. Hence the sum of each position, whether units, 10s or otherwise is going to be (2+3+4+5+6 = 20)
Hence we have (10000)(20)(5!) + (1000)(20)(5!)+ (100)(20)(5!) + (10)(20)(5!) + (1)(20)(5!)
= (20)(5!)(10000 + 1000 + 100+ 10+ 1) = (20)(5!)(11111)
What i cant understand is how they arrived at 4! instead of 5!.
Please help !!
The solution given is (2+3+4+5+6)*(11111)*4!
Why is it 4! ??
I think the solution is (2+3+4+5+6)*(11111)* 5!
The number of ways in which 5 digits can be arranged into 5 digit numbers without repetition is 5!.
Each digit will repeat in each of the positions once. Hence the sum of each position, whether units, 10s or otherwise is going to be (2+3+4+5+6 = 20)
Hence we have (10000)(20)(5!) + (1000)(20)(5!)+ (100)(20)(5!) + (10)(20)(5!) + (1)(20)(5!)
= (20)(5!)(10000 + 1000 + 100+ 10+ 1) = (20)(5!)(11111)
What i cant understand is how they arrived at 4! instead of 5!.
Please help !!
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28 Mar 2011, 02:28
Kudos
Bookmarks
Consider the sum of just the units digit of all such numbers.
No. of numbers ending in 2 = 4!
No. of numbers ending in 3 = 4!
No. of numbers ending in 4 = 4!
No. of numbers ending in 5 = 4!
No. of numbers ending in 6 = 4!
So sum of units digit = (2+3+4+5+6)*1*4!
Similarly for the sum of just the 10's place
No. of numbers with 2 in ten's place = 4!
No. of numbers with 3 in ten's place = 4!
No. of numbers with 4 in ten's place = 4!
No. of numbers with 5 in ten's place = 4!
No. of numbers with 6 in ten's place = 4
So sum of tens digit = (2+3+4+5+6)*10*4!
....
You continue with that logic to get :
(2+3+4+5+6)*(1+10+100+1000+10000)*4! = (2+3+4+5+6)*11111*4!
No. of numbers ending in 2 = 4!
No. of numbers ending in 3 = 4!
No. of numbers ending in 4 = 4!
No. of numbers ending in 5 = 4!
No. of numbers ending in 6 = 4!
So sum of units digit = (2+3+4+5+6)*1*4!
Similarly for the sum of just the 10's place
No. of numbers with 2 in ten's place = 4!
No. of numbers with 3 in ten's place = 4!
No. of numbers with 4 in ten's place = 4!
No. of numbers with 5 in ten's place = 4!
No. of numbers with 6 in ten's place = 4
So sum of tens digit = (2+3+4+5+6)*10*4!
....
You continue with that logic to get :
(2+3+4+5+6)*(1+10+100+1000+10000)*4! = (2+3+4+5+6)*11111*4!
28 Mar 2011, 02:40
Kudos
Bookmarks
amoghpanje
Let us first confirm what is the right answer:
2,3,4,5,6 can be arranged in 5! ways
23456
23465
23564
.
.
.
______
ADDITION
If there are 5!=120 numbers in total. The unit digits of all numbers must be equally distributed. None of the digits will have preference over others.
Thus, in all of these 5!=120 numbers, at units place 2 will appear (120/5=24) times, 3 will appear 24 times, 4,5,6 will also appear 24 times each;
Let's find the sum of the units digits of all 120 numbers:
24*2+24*3+24*4+24*5+24*6 = 24*(2+3+4+5+6) = 24*20= 480
0 at Unit's digit of the result and 48 carry over
Let's find the sum of the tens digits of all 120 numbers. It is not going to be any different.
24*2+24*3+24*4+24*5+24*6 = 24*(2+3+4+5+6) = 24*20= 480+48(Carried over from unit) = 528
8 at Tens digit of the result and 52 carry over
Likewise;
480+52=532
2 at Hundreds digit of the result and 53 carry over
480+53=533
3 at Thousands digit of the result and 53 carry over
480+53=533
533 will be the first 3 digits of the result.
Result of Addition = 5333280
(2+3+4+5+6)*(11111)*4! = 5333280 (Matches)
(2+3+4+5+6)*(11111)* 5! = 26666400(Doesn't match)
If you see the red part in your question, the multiplication by 5! will mean that the set (2+3+4+5+6) appear in 120 numbers 120 times whereas we should multiply it with 24=4! as this set (2+3+4+5+6) appear in 120 numbers 24 times for every place(unit, tens, etc.).
