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.
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, nPr = nCrr!
? nPr = 720nCr
? nCr.r! = 720nCr
? r! = 720
? r = 6
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
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
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
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
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