If there were no three points collinear. We should have ¹⁰C₂ lines but since 7 points are collinear we must subtract ⁷C₂ lines and add the one corresponding to the line of collinearity of the seven points.

Correct Option: C

If there were no three points collinear. We should have ¹⁰C₂ lines but since 7 points are collinear we must subtract ⁷C₂ lines and add the one corresponding to the line of collinearity of the seven points.
Thus, the required number of straight lines .
= ¹⁰C₂ - ⁷C₂ + 1 = 25

Required no. of ways = [⁴C₁ x ⁶C₃] + [⁴C₂ x ⁶C₂] + [⁴C₃ x ⁶C₁] + [⁴C₄]
= 80 + 90 + 24 + 1 = 195.

Correct Option: C

Required no. of ways = [⁴C₁ x ⁶C₃] + [⁴C₂ x ⁶C₂] + [⁴C₃ x ⁶C₁] + [⁴C₄]
= 80 + 90 + 24 + 1 = 195.

We may choose 1 officer and 5 jawans or 2 officers and 4 jawans ......... or 4 officers and 2 jawans.

Correct Option: A

We may choose 1 officer and 5 jawans or 2 officers and 4 jawans ......... or 4 officers and 2 jawans.

So Required answer = [⁴C₁ x ⁸C₅] + [⁴C₂ x ⁸C₄] + [⁴C₃ x ⁸C₃] + [⁴C₄ x ⁸C₂]
= 224 + 420 + 224 + 28 = 896.

Total number of letters = 5
Number of vowels = 2.
If we consider both vowel as a one letter then,
Required number = 4! 2! = 48.

Correct Option: A

Total number of letters = 5
Number of vowels = 2.
If we consider both vowel as a one letter then,
Required number = 4! 2! = 48.

The number of ways of selecting 3 points out of 12 points is ¹²C₃. The number of ways of selecting 3 points out of 7 points, on the same straight line is ⁷C₃

Correct Option: C

The number of ways of selecting 3 points out of 12 points is ¹²C₃. The number of ways of selecting 3 points out of 7 points, on the same straight line is ⁷C₃, Hence, the number of triangle formed will be ¹²C₃ - ⁷C₃ = 210 - 35 = 185.