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Question

The number of different seven digit numbers that can be written using only three digits 1,2 & 3 under the condition that the digit 2 occurs exactly twice in each number is

A
672
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B
640
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C
512
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D
none of these
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Solution

The correct option is C 672
Total number of digits =37
The 7 digit number has exactly two digits 2
Let's say the number
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯22cdefg,¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯2b2defg,¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯2bc2efg,..
Number of cases we can put exactly two digits (out of 7 digits) are 2 is
6+5+4+3+2+1=21cases.
For each case, the remaining 5 digits can be either 1 or 3, so the number of ways for each case is
25=32 ways
the number of favorable numbers is
=21×32=672 numbers

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