in Graph Theory edited by
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https://gateoverflow.in/931/gate2003-40

IN THE SOLUTION AT LAST 3 STATEMENTS

0≤−6.

which is definitely inconsistent.

Hence, answer = option (D)

WE PUT 6 HERE THEN IF WE DONT HAVE VALUES MEANS FILL IN THE BLANKS THEN..............

IS THERE ANY METHOD TO SOLVE THIS OR ANYTHING ??????

 

 

 

in Graph Theory edited by
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let assume, minimum degree = 6

===>$\frac{min . |v| }{2} $ ≤ 3.|v| - 6

leads to $\frac{6 . |v| }{2} $ ≤ 3.|v| - 6 ===> 3|v| ≤ 3.|v| - 6 ===> 3|v| - 3|v| ≤ 6 ===> 0 ≤ 6 which is wrong

therefore it can't be 6

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@Shaik Masthan ITS JUST HIT AND TRAIL METHOD . I AM GETTING IT.

BUT IF IT IS IN THE FILL IN THE BLANKS THEN .................

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edited by
$\frac{min . |v|}{2}$ ≤  3|v| - 6 ===> min . |v| ≤  6|v| - 12 ===> min ≤  $\frac{6|v|-12}{|v|}$ ===> min ≤  $\frac{6|v|}{|v|} - \frac{12}{|v|}$ ===> min ≤  $6 - \frac{12}{|v|}$,

note that it is should be lessthan 6 ===> 5.___

===> min can't be 6
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