Given statements
Statement I: The number of points (x, y, z) in space, whose each co-ordinate is a negative integer such that x + y + z + 12 = 0 is 55
Statement II: The number of ways in which 9 identical balls can be placed in three identical boxes is 6
Statement III: If $x_{1} x_{2} x_{3} = 2 \times 5 \times 7^{2},$ then the number of solution set for $(x_{1}, x_{2}, x_{3})$ where $x_{i} \in N, x_{i} > 1$ is 21
Statement IV: The number of 4 letter words that can be formed using the letters of DEEPMALA is 600
Statement V: Number of ordered triplets of natural numbers (a, b, c) for which abc $\leq$ 11 is 56
Statement VI: Let $\bar{r}$ be a variable vector and $\bar{a} = \hat{i} + \hat{j} + \hat{k}$ such that the scalars $(\bar{r} . \hat{i}), (\bar{r} . \hat{j}), (\bar{r} . \hat{k})$ are positive integers. If $\bar{r} . \bar{a}$ is not greater than 10, then total number of possible vectors $\bar{r}$ is 119
Which of the following is/are CORRECT?

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I,III, V are correct

II,IV, VI are false

I,IV,VI are correct

II,IV,V are false

answer is A, B.

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Detailed Solution

ST I: x + y + z = -12, x < 0, y < 0, z < 0
- x > 0, - y > 0, -z > 0 (or) –x > 1, -y > 1, -z > 1
Let $- x = α, - y = β, - z = γ$

$α + β + γ = 12, α \geq 1, β \geq 1, γ \geq 1$

ST II: Required number of ways = $(3 c_{1} x 2 c_{1} x 1 c_{1}) 2 = 12$
ST III: $x_{1} x_{2} x_{3} = 2 x 35 x 7 = 2 x 49 x 5$
$= 10 x 7 x 7 = 14 x 7 x 5$
no. of solution set ix $3 \times 3! + \frac{3!}{2!}$
ST IV: R.N.W = $^{6} C_{4} \times 4! + 2 (^{5} C_{2} \times \frac{4!}{2!}) + 1 \times \frac{4!}{2! 2!} = 360 + 240 + 6 = 606$
ST V: abc = 1 in 1 way
abc = 2,3,5,7,11 in 15 ways
abc = 4,9, in 12 ways
abc = 8 in 10ways
abc =6,10 in 18 ways
Total = 56
ST VI: $x + y + z \leq 10 \Rightarrow R . N . W =^{10} C_{3}$