how many three-digit positive integers have three different odd digits?

Question

please show work because i don't get it. i would appreciate any help

Answer 1

Take the odd digits {1,3,5,7,9}

You have 5 choices for the first digit.
You have 4 choices for the second digit.
You have 3 choices for the third digit.

5 x 4 x 3 = 60

60 ways to have a three-digit positive integer with different odd digits.

Answer 2

There are 60 nos. in total with three different digits in the three digit odd integers

Consider 3 digits starts with 100 and end in 999

Only the 100s, 300s, 500s, 700s & 900s to be considered, as these are starting with odd nos. 200 and its multiples will have atleast one even nos between all three digits.

In 100 to199,

Consider the 10th digit. 110s, 120s, 130s, 150s up to 190s. There are 120s, 140s 160s &180s having the even integers. Consider only 130s, 150s, 170s, 190s. (Don't consider 110s as it has two odd integer).

130s has, 135, 137, 139 simillarly
150s having 153, 157, 159,
170s having 173, 175, 179,
190s having 193, 195, 197


So for a three digit integer between 100 and 199, the total number of the three different integers having all three digits as odd digits is 12

So for the rest 300s, 500s, 700s & 900s will have each 12 nos.,

(compare the result, if you can, by writing or checking it from any number table, it is correct)

Consider all the 5 cases and 12 nos. from each case gives you,

5*12=60 nos.

Answer 3

Since 135, 315, etc. are different three-digit integers we must consider all permutations of 1,3,5,7,9 taken three at a time.

Since you haved 5 choices for the first odd digit, 4 for the second and 3 for the third there will be

5 x 4 x 3 = 60 different 3 digit positive integers with
different odd digits.

Answer 4

5*4*3 = 60

100's digit: 5 options to choose from (1, 3, 5, 7, 9)
10's digit: 4 options to choose from (whatever's not used for 100's digit)
1's digit: 3 options to choose from (whatever's left over)

Answer 5

there are 5 odd digits 1,3,5,7,9 we want to put them in 3 places without repetiton with respect to position so use the rule
n! / (n-r)!
so=5!/(5-3)! = 1*2*3*4*5 / (1*2) =3*4*5 = 60