If 7 points out of 12 are in the same straight line, then the number of triangles formed is?

The hundred place will be reserved for 3 or 2
5 digits are free to fill rest two places i,e., of tens and unit.

Number of required 3 digit number = 2 x ⁵C₂ = 20

We know that, ⁿP_r = ⁿC_rr!
? ⁿP_r = 720ⁿC_r
? ⁿC_r.r! = 720ⁿC_r
? r! = 720
? r = 6

We have If ^{2n + 1}P_{n - 1} / ^{2n - 1}P_n = 3/5
? 5 x ^{2n + 1}P_{n - 1} = 3 x ^{2n - 1}P_n
? {5 x (2n + 1)!} / {(n + 2 )!} = 3 x {(2n - 1)!} / {(n - 1)!}
? 10(2n + 1) = 3(n + 2) (n + 1)
? 3n² - 11n - 4 = 0
?(3n + 1 ) (n - 4) = 0
? n = 4

³⁵C_{n + 7} = ³⁵C_{4n - 2}
We know that if ⁿC_x = ⁿC_y then, Either x = y or x + y = n

Case 1: x + y = n
? n + 7 + 4n - 2 = 35
? 5n + 5 = 35
? 5n = 30
? n = 6

Case 2: x = y
? n + 7 = 4n - 2
? 4n - n = 2 + 7
? 3n = 9
? n = 3

ⁿC_{r - 1} = 36, ⁿC_r = 84 and ⁿC_{r + 1} = 126
? ⁿC_r / ⁿC_{r - 1} = 84/36
? 3n - 10r + 3 = 0 ....(i)

ⁿC_{r + 1} / ⁿC_r = 126/84
? 2n - 5r - 3 = 0 .............(ii)

From (i) and (ii) r = 3, n = 9

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

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.

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

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

There are 3A's 2N's and one B. We have to find the total number of arrangements of 6 letters out of which 3 are alike of one kind, 2 are alike of second kind, thus the total number of words
= 6! / (3! 2!) = 60