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

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{ How many different 7 digit members are their sum }\hspace{.33em}\\~\\ & \normalsize \text{ of whose digits are odd ?}\hspace{.33em}\\~\\ & a.)\ 45\times 10^{5} \hspace{.33em}\\~\\ & b.)\ 24\times 10^{5} \hspace{.33em}\\~\\ & c.)\ 224320 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{ none of these } \hspace{.33em}\\~\\ \end{align}}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1odd + odd = even even + even = even odd + even = odd we need an odd number of odd digits

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1only 1 odd, 3 odd, 5 odd are allowed

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & 5\ odd +2\ even \hspace{.33em}\\~\\ & 3\ odd +4\ even \hspace{.33em}\\~\\ & 1\ odd +6\ even \hspace{.33em}\\~\\ \end{align}}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1first 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 \(5^7\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1now 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 oddnumber 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 \(\binom{6}{5}\times 4 \times 5^6\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The answer is approximately A.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[44.21875 \times 10^5\]that's more like what I get.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Maybe I've not counted something, because the answer is exactly A.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is this true there are 9 * 10^6 seven digit numbers and half of them have sum of digits odd

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2@parthkohli If we count 7 odd numbers which gives 5^7=78125. 4421875+78125=4500000, bingo!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1that's the stupidest mistake i could have made

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1we can also prove that there's a bijection from the set of odd sumofdigits to even sumofdigits to prove that the cardinality of the two sets is the same.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Then we need to accept 7 digit numbers with leading zeroes!
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