statistic trouble help

[jinx0007]

Badly need help to attempt this question because i am 90% sure a question like this will appear in the exams that i will be taking next week.

Find out how many 3 digit odd number that can be obtained form digits 1, 2, 3, 4, 5, 6, 7 if

1/ repetition of digit not allowed

2/ Repetition of digit allowed

ANSWER: 1/ 120 2/ 196


MY ATTEMPT

1/

- - - I'VE CONTROLLED THE LAST BLOCK 1, 3, 5, 7

6p2 X 4 = 120

2/

--- I'VE CONTROLLED THE LAST BLOCK 1, 3, 5, 7

AS DIGIT REPEAT

7P2 X 4 = 148

i AM A BIT STUCK IN THE SECOND PART BUT PLEASE VERIFY THE FIRST PART AND HELP ME TO ATTEMPT THE SECOND PART

 

[AuTh3nTiK]

for the first part i do it like this:
___ ___ ___
6c1 x 5c1 x 4c1
you can only use the digits 1,3,5 or 7 so the last block is 4c1
then you you will be left with 6 digits. so you start with the first block which will be 6c1 and the second block 5c1. You multiply it and you will get the ans as 120

Second part:

___ ___ ___
7c1 x 7c1 x 4c1
the last block can only contain odd so it is 4c1
since digits may be repeated you may use all digits again so with both remaining blocks you get 7c1
you multiply and get ans=120

hope you understand and best of luck for the exams...

 

[jinx0007]

hmmm thnks but i dont think that the answer is correct...simple logic: whose with repetition and those with not cannot have the same answer that is 120 ways....infact the answer for the second part is 196 according to the booklet..!!!!

any help are welcome THNKS..!!!

 

[jinx0007]

awww no the answer is correct lolz...sorry...i have mistype the figures in my calculator...thanks dude..!!!

 

[AuTh3nTiK]

yeah the last part i have mistaken it the is 196...sorry