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
Total number of books = a + 2b + 3c + d . Since there are 'b' copies of each of two books, 'c' copies of each of three books and single copy of 'd' book.
Therefore, the total number of arrangements is = (a + 2b + 3c + d )! / {a! (b!)2 (c!)3}
Required number of lines = nC2
= 15C2
= (15 x 14)/2
= 105
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.
nCr - 1 = 36, nCr = 84 and nCr + 1 = 126
? nCr / nCr - 1 = 84/36
? 3n - 10r + 3 = 0 ....(i)
nCr + 1 / nCr = 126/84
? 2n - 5r - 3 = 0 .............(ii)
From (i) and (ii) r = 3, n = 9
35Cn + 7 = 35C4n - 2
We know that if nCx = nCy 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 + 1Pn - 1 / 2n - 1Pn = 3/5
? 5 x 2n + 1Pn - 1 = 3 x 2n - 1Pn
? {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, nPr = nCrr!
? nPr = 720nCr
? nCr.r! = 720nCr
? r! = 720
? r = 6
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