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  • The digits 1,1, 2, 2, 2, 5, 3, 6, 6, 6 are used to create 10-digit numerals. The number of such numbers that do not have odd digits in odd places is  <s

The digits 1,1, 2, 2, 2, 5, 3, 6, 6, 6 are used to create 10-digit numerals. The number of such numbers that do not have odd digits in odd places is

 

Option: 1

1200


Option: 2

600


Option: 3

500


Option: 4

1800


Answers (1)

best_answer

Given that,

We must arrange the given numbers so that none of the eight-digit number's odd positions are taken up by odd numbers.

Therefore, out of those mentioned numbers, 1, 1, 5, and 3 are the only 4 odd numbers.

Additionally, there are 4 even numbers: 2, 2, 2, 6, 6, and 6.

And as we know that there are 5 odd places in a 10-digit number.

We have to fill the odd places with even numbers and then fill the left even places with other numbers

So, for that we have to choose 5 numbers out of 6 given even numbers.

There are only 5 odd places.

And then fill all even places with 4 odd numbers and one even number that is left.

So, for doing this there can be two cases.

Case 1: odd places filled by (2, 2, 2, 6, 6) and even places filled by (1,1,5,3, 6)

Thus, the number of ways is given by,

\begin{aligned} & N=\left(\left(\frac{5 !}{3 ! 2 !}\right) \text { for odd places }\right) \times\left(\left(\frac{5 !}{2 !}\right) \text { for even places }\right) \\ & N=10 \times 60 \\ & N=600 \end{aligned}

Case 2: odd places filled by (2, 2, 6, 6, 6) and even places filled by (1,1,5,3, 2)

Thus, the number ways is given by,

\begin{aligned} & N=\left(\left(\frac{5 !}{3 ! 2 !}\right) \text { for odd places }\right) \times\left(\left(\frac{5 !}{2 !}\right) \text { for even places }\right) \\ & N=10 \times 60 \\ & N=600 \end{aligned}

Therefore, the total number of ways will be 600 + 600 = 1200.

 

Posted by

Pankaj Sanodiya

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