JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 1)
Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :
$${5 \over 2}\left( {6!} \right)$$
$${6!}$$
56
$${1 \over 2}\left( {6!} \right)$$
Explanation
Here none number repeats more than once.
We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in 5C1 ways.
Let number 3 repeats more than once. So six digits are 1, 3, 3, 5, 7, 9.
We can arrange those six digits in $${{6!} \over {2!}}$$ ways.
$$ \therefore $$ Total six digit numbers = 5C1 $$ \times $$ $${{6!} \over {2!}}$$ = $${5 \over 2}\left( {6!} \right)$$
We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in 5C1 ways.
Let number 3 repeats more than once. So six digits are 1, 3, 3, 5, 7, 9.
We can arrange those six digits in $${{6!} \over {2!}}$$ ways.
$$ \therefore $$ Total six digit numbers = 5C1 $$ \times $$ $${{6!} \over {2!}}$$ = $${5 \over 2}\left( {6!} \right)$$
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