If ⁿC _{r - 1} = 36, ⁿC _r = 84 and ⁿC _{r + 1} = 126 then?

We have 9 letters 3 a's, 2 b's and 4 c's. These 9 letters can be arranged in 9! / (3! 2! 4!) = 1260 ways.

Required number of ways = (2 + 1) (3 + 1) (4 + 1) - 1 = 59

12 persons can be seated around a round table in 11! ways. The total number of ways in which 2 particular persons sit side by side = 10! x 2!.

Hence, the required number of arrangements = 11! - 10! x 2! = 9 x (10!)

Required no. of words = ¹²P₄ x ⁴P₃
= 12 x 11 x 10 x 9 x 4 x 3 x 2
= 120(100 - 1) x 24
= 288000 - 2880
= 285120

Required no. of triangles = ¹²C₃ - ⁷C₃ = 185

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

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

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

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

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.