The numbers of straight lines can be formed out of 10 points of which 7 are collinear?

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.

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.

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

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

Let there be n persons in the room. The total number of hand shakes is same as the number of ways of selecting 2 out of n.
ⁿC₂ = 66
? n(n - 1) / 2! = 66
? n² - n - 132 = 0
? (n - 12) (n + 11) = 0
? n = 12

There are 12 letters in the world 'civilization' of which four are i's and other are different letters.
? Total number of permutations = 12!/4!
But one word is civilization itself.
? Required number of rearrangements = 12!/4! - 1

Taping all vowels (IEO) as a single letter (since they come together ) there are six letters with two 'R' s
Hence no. of arrangement = 6!/2! x 3! = 2160
[3 vowels can be arranged in 3! ways among themselves, here multiplied with 3!. ]

Possible arrangements are 7!/2! 2! = 1260
[Division by 2 times 2! is because of the repletion of E and R .]