how many different numbers can be made with the digits 1,3,5,7,9 when taken all at a time,and what is there sum?
2007-08-10 04:23:31 · 8 answers · asked by sameerteacher01 1 in Science & Mathematics ➔ Mathematics
Different numbers:
You have 5 numbers, so the permutations are:
5*4*3*2*1 = 120
The sum would be:
1*(1 + 3 + 5 + 7 + 9)
plus
10*(1 + 3+ 5+ 7 + 9)
plus
100*(1 + 3 + 5 + 7 + 9)
plus
1000*(1 + 3 + 5 + 7 + 9)
plus
10000*(1 + 3 + 5 + 7 + 9)
Then all of that times 120/5. (It's 120/5 because the above sum added 5 of those permutations.)
so...
1+3+5+7+9 = 25
so...
(120/5)(25 + 250 + 2500 + 25000 + 250000)
= 24(277775)
= 6,666,600
2007-08-10 04:30:38 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
You can keep either of digits in 10thousand's place
i.e you can put 5 digits in 10 thouasnd's place.
Similarly, you can put 4 digits in thouasnd's place.
you can put 3 digits in hundred's place.
you can put 2 digits in ten's place.
you can put 1 digits in unit's place.
So total numbers formed are,
5! = 5*4*3*2*1 = 120
Total numbers = 120 Digits used are 5 hence each digit occures 120/5 = 24 times.
The sum =
24Ã (1+3+5+7+9)Ã10000 + 24Ã (1+3+5+7+9)Ã1000 + 24Ã (1+3+5+7+9)Ã100 + 24Ã (1+3+5+7+9)Ã10 + 24Ã (1+3+5+7+9)Ã1
24 (250000 + 25000 + 2500 + 250 + 25)
= 24(277775)
= 6,666,600
2007-08-12 01:31:37 · answer #2 · answered by Pranil 7 · 0⤊ 0⤋
Different no.s would be :
You have 5 numbers, so the permutations are:
5! = 5*4*3*2*1 = 120
The sum would be :
Here...
1+3+5+7+9 = 25
so...
(120/5)(25 + 250 + 2500 + 25000 + 250000)
= 24(277775)
= 6,666,600 [ans]
2007-08-10 07:32:26 · answer #3 · answered by sharbadeb 2 · 0⤊ 0⤋
There are 5*4*3*2*1= 120 possible numbers with
digits 1,3,5,7, 9 when taken all at a time.
As for the sum, Mathematica has it spot on.
2007-08-10 04:51:33 · answer #4 · answered by steiner1745 7 · 0⤊ 0⤋
There are 5 ways to choose a first digit, 4 ways to choose a (different) second digit, 3 ways to choose a third, 2 ways to choose a fourth, and just 1 way to choose a last. Therefore, there are 5*4*3*2*1 = 5! = 120 ways.
2007-08-10 04:33:03 · answer #5 · answered by Tony 7 · 0⤊ 0⤋
1,3,5,7,9
how many different numbers = 5! = 120 numbers
what is there sum = 6 666 600
this is a correct ans.
2007-08-10 04:27:28 · answer #6 · answered by harry m 6 · 0⤊ 0⤋
5!= 5x4x3x2x1=120
25
25
25
25
25
then their sum =(120/5 )277775=24x277775=6666600
2007-08-10 04:41:40 · answer #7 · answered by mramahmedmram 3 · 0⤊ 0⤋
for the first part, multiply 5*4*3*2*1, which is 120. that means you can combine them and get 120 different numbers.
I don't know about the sum part.
2007-08-10 04:38:16 · answer #8 · answered by lizie 4 · 0⤊ 0⤋