Permutation and combination homework

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The discussion focuses on calculating the number of three-digit odd numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, and 7, with and without repetition of digits. For the case without repetition, the last digit must be odd (1, 3, 5, or 7), leading to a calculation of 120 valid combinations. The user expresses confusion regarding the method for the second case, where repetition is allowed. The correct answer for this scenario is 196, which involves determining choices for each digit position after selecting the last digit. The conversation highlights the importance of understanding the selection process for each digit in forming valid numbers.
jinx007
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Find how many 3 digits odd number that can be obtained from the digit 1,2,3,4,5,6,7 if,

1/ Repetition of digits not allowed
2/ Repetition of digits allowed

my work

1/ --- the last i digit i have control 1,3,5,7

so 2 remaining digits = 6 P 2 x 4 (4ways)

= 120

2/ i AM STUCK

In fact, i don't really know whether the method is correct,i am a bit confuse, HELPPPP

ANSWER ACCORDING TO BOOKLET :
1/ 120

2/ 196
 
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For the second question, how many choices for the rightmost digit do you have?
After that, how many choices for the middle digit? And the first? So...
 
nvm i gave away too much for a homework question
 

Thread 'Family of lines that are at a distance of 5 from the origin'

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