 In constructing a problem on vectors, the three components of a vector are randomly chosen from the digit 0 to 5 with replacements. The probability that the magnitude of the vector is 5, is(A) $\dfrac{1}{{24}}$(B) $\dfrac{1}{{12}}$(C) $\dfrac{1}{6}$(D) $\dfrac{1}{{30}}$ Verified
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Hint: This is the question of probability but we have to have knowledge of vectors also. The basic concept that we use in this is we make all possible combinations of the component whose magnitude will be 5. A vector can be represented by $xi + yj + zk$ form so we make a pair of 3 digits whose magnitude is 5.

Given that we have total 6 digit that is $0,1,2,3,4,5$
And in a vector $xi + yj + zk$
$x, y, z$ are filled by all 6 digit given above
So we can fill $x$ by 6 number of ways
$y$ can also fill by 6 number of ways
Similarly $z$ is also fill by 6 number of ways
So total number of possible combination is = $6 \times 6 \times 6$
So total number of possible combination is = $216$
Now the possible combination of the components whose magnitude will be 5 is : $(0,0,5)$ and $(0,3,4)$ these are only possible combinations.
Now number of arrangement of $(0,0,5)$ = $\dfrac{{3!}}{{2!}}$ ( 2 number are same for that unlike arrangements is $2!$)
Number of arrangements of $(0,3,4)$ = $3!$
So total number of such combination is $\dfrac{{3!}}{{2!}} + 3!$
$\Rightarrow 3 + 6 = 9$
So total number of combination of : $(0,0,5)$ and $(0,3,4)$ is $9$
Now probability is found by dividing by combination of $(0,0,5)$ and $(0,3,4)$ and total number of combinations.
Hence, probability = $\dfrac{9}{{216}}$
Probability = $\dfrac{1}{{24}}$

So, the correct answer is “Option A”.

Note: the formula for a combination of choosing $r$ unique ways from $n$ possibilities is: ${}^nC{}_r = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
where $n$ is the number of items and $r$ is the unique arrangements.
And the formula for the combination of $n$ possibilities but out of them $r$ is the same as $\dfrac{{n!}}{{r!}}$.