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.

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

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