## mathmath333 one year ago How many different 7 digit members are their sum of whose digits are odd ?

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{ How many different 7 digit members are their sum }\hspace{.33em}\\~\\ & \normalsize \text{ of whose digits are odd ?}\hspace{.33em}\\~\\ & a.)\ 45\times 10^{5} \hspace{.33em}\\~\\ & b.)\ 24\times 10^{5} \hspace{.33em}\\~\\ & c.)\ 224320 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{ none of these } \hspace{.33em}\\~\\ \end{align}}

2. ParthKohli

odd + odd = even even + even = even odd + even = odd we need an odd number of odd digits

3. ParthKohli

only 1 odd, 3 odd, 5 odd are allowed

4. mathmath333

\large \color{black}{\begin{align} & 5\ odd +2\ even \hspace{.33em}\\~\\ & 3\ odd +4\ even \hspace{.33em}\\~\\ & 1\ odd +6\ even \hspace{.33em}\\~\\ \end{align}}

5. ParthKohli

first consider the cases where the first digit is taken by an odd number... (i) 1 odd number O _ _ _ _ _ _ the only odd number is in the first digit. 1, 3, 5, 7, 9 are the possible choices. the rest of the digits are even. so 0, 2, 4, 6, 8 are the choices for those. overall, $$5^7$$ numbers in this category (ii) 3 odd numbers O _ _ _ _ _ _ first we choose the two remaining places in $$\binom{6}{2}$$ ways. now we have $$5$$ choices for each of those three places for odd numbers and $$5$$ choices for all other places too. so the count is $$\binom{6}{2}\times 5^7$$ for this category. (iii) 5 odd numbers O _ _ _ _ _ _ first choose the four remaining places in $$\binom{6}{4}$$ ways, then multiply by $$5^7$$

6. ParthKohli

now consider the cases where an odd number does NOT occupy the first place E _ _ _ _ _ _ the possible choices for the first digit are 2, 4, 6, 8 and the rest of the even digits are 0, 2, 4, 6, 8. so then (i) one odd number E _ _ _ _ _ _ choose the place for the odd-number in $$\binom{6}{1}$$ ways first. for the first place, 4 choices. all other six places - five choices. so $$\binom{6}{1} \times 4 \times 5^6$$ ways. (ii) three odd numbers E _ _ _ _ _ _ $$\binom{6}{3}$$ ways to choose the place. now the count is $$\binom{6}{3}\times 4 \times 5^6$$ (iii) five odd numbers E _ _ _ _ _ _ $$\binom{6}{5}$$ ways to choose the place. now the count is $$\binom{6}{5}\times 4 \times 5^6$$

7. ParthKohli

8. ParthKohli

$44.21875 \times 10^5$that's more like what I get.

9. ParthKohli

Maybe I've not counted something, because the answer is exactly A.

10. mathmath333

is this true there are 9 * 10^6 seven digit numbers and half of them have sum of digits odd

11. mathmate

@parthkohli If we count 7 odd numbers which gives 5^7=78125. 4421875+78125=4500000, bingo!

12. ParthKohli

that's the stupidest mistake i could have made

13. ParthKohli

we can also prove that there's a bijection from the set of odd sum-of-digits to even sum-of-digits to prove that the cardinality of the two sets is the same.

14. mathmate

Then we need to accept 7 digit numbers with leading zeroes!

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