How many 9 digits numbers can be formed from the digits of the number 223355888 by arrangement of the digits so that the odd digit occupy even places?

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Solution

Odd digit- $3355$
Even digit - $22888$
There are $4$ even placed and $5$ odd places
Odd Number are placed in $= \frac{4!}{2! 2!}$ ways
$= 6$
Even Numbers are placed in $= \frac{5!}{2! 3!} = 10$ ways
$∴$ No. of required of digit number $= 6 \times 10 = 60$ ways

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How many 9 digits numbers can be formed from the digits of the number 223355888 by arrangement of the digits so that the odd digit occupy even places?

Odd digit- 3355Even digit - 22888There are 4 even placed and 5 odd placesOdd Number are placed in =4!2! 2! ways=6Even Numbers are placed in =5!2! 3!=10 ways∴ No. of required of digit number =6×10=60 ways

Odd digit- $3355$
Even digit - $22888$
There are $4$ even placed and $5$ odd places
Odd Number are placed in $= \frac{4!}{2! 2!}$ ways
$= 6$
Even Numbers are placed in $= \frac{5!}{2! 3!} = 10$ ways
$∴$ No. of required of digit number $= 6 \times 10 = 60$ ways