How many numbers can be formed using all the digits 1 6 7 8 6 1 so that odd digits always occuppy odd positions?
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Final Answer : 9
Steps:
1) There are 3 odd and 3 even places in 6 - digit no.
Odd numbers are 1,1,7
Odd numbers can be arranged in 3!/2! = 3ways.
2) Even numbers are 6,8,6
Even numbers can be arranged in 3!/2! = 3 ways.
So, Required no. of ways = 3* 3 = 9ways.
Steps:
1) There are 3 odd and 3 even places in 6 - digit no.
Odd numbers are 1,1,7
Odd numbers can be arranged in 3!/2! = 3ways.
2) Even numbers are 6,8,6
Even numbers can be arranged in 3!/2! = 3 ways.
So, Required no. of ways = 3* 3 = 9ways.
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