In how many possible ways can you write 1800 as a product of 3 positive integers a, b, c ?

Prime Factorization of 1800 
2^3 * 3^2 * 5^2

Distribute power of prime factors among digits a, b,c as asked. For example
If we distribute 2^4 as (2,2,0) and, lets say a,b,c with powers of other prime factors 3 and 5 are (1, 9,25) ,then, with powers of 2 distributed as (2,2,0) (total 4), becomes
(2^2 *1, 2^2 * 9, 2^0 * 25) = (4, 36,25 )= 4*36*25 = 1800
Take another example
Distribute as (0,3,1)
(2^0 * 1, 2^3 * 9, 2^1 * 25) = (1, 72,25) = 1800
2^4 can be distibuted like the followings among 3 numbers (a,b,c). Smilarly powers of other prime factors. Together the product is 1800
4,0,0
0,4,0
0,0,4
3,1,0
3,0,1
1,3,0
1,0,3
......
2,2,0
2,0,2
0,0,2
2,1,1
1,2,1
1,1,2
...etc. etc
So problem becomes distributing power r (4 here for prime factor 2) among n (3 here) numbers.
 Distribute r items among n people
D_(r,n)=\frac{(r+ (n1))!}{r! (n1)!}
Find this sequence of combinations D_(r,n)
in Pascals triangle
Answer is D_{(3,3)} * D_{(2,3)} * D_{(2,3)} = 10 * 6 * 6
Ans: 360
How many ways letters in word ARMOUR can be arranged
n = 6
repeating letters = R 2 times, Combinations of them are indistinguishable so will have to reduce from total permutations
So net permutations = \frac{n!}{r!}
= 720 / 2 = 360
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?
Number of ways 3 consonants and 2 vowels \ can be selected =
here combination is taken because order of letters not important. However order is important when you form words with these selected letters
=\frac{7!}{(73)! 3!} * \frac{4!}{(42)! 2!} \\
=35 * 6\\
= 210 \\
Permutations of words with 3 consonants and 2 vowels
=5!
= 120
So total combinations = 120 * 210 = \underline{25200}
How many 3digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated ?
Divisible by 5 means last digit is 5.
For remaining 2 digits, total permutations without repeat =
= {^5\!P_2}
=\underline20
Evaluate permutation equation 59P3 ?
= 59 X 58 X 57
How many words can be formed from the letters of the word 'AFTER', so that the vowels never come together ?
For 5 letters, total permutations 5!
Consider A and E as one letter , ie. total 4 letters and then permutations would be 4!
AE can have two arrangements, AE and EA= 2!
So total permutations with AE and EA considered one letter = 2 * 4!
So total permutations where A and E doesn't come together
= 5!  2*4! = 72
A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colors ?
Since there are three colors and three draws, different color should come in each draw, so total number ways three different colors can be draws in three draws = 4 3 2 = 24
In how many different ways the letters of the word "MATHEMATICS" can be arranged so that the vowels always come together ?
Total 11 letters
Vowels  AEAI. A repeats
Consonants M and T repeats
Consider vowels AEAI as one letters =V, since it is asked to find arrangements where they come together. However A repeats so, combination needs to divide by 2!
Total number of letters with AEAI together as one letter = 8. Total arrangements = 8!. However M and T repeats . So arrangements become 8!/(2!*2!)
Now with AEAI, they can have their own arrangements, still being together. So those arrangements are 4!/2! (since A repeats)
total permutations= 8! x 4!/8=7! * 4! = 120960
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw ?
Total Combinations of 9 balls into 3= {}^9C_3
Total combination where black ball doesn't show up = {}^{(number of other balls)}C_3 = []^6C_3
Number of combinations where at least one black ball shows up = {}^9C_3  {}^6C_3
= 8420 = 64
How many groups of 6 persons can be formed from 8 men and 7 women ?
{}^{15}C_3
8 men entered a lounge simultaneously. If each person shakes hand with the other, then find the total no. of handshakes?
shake hand is group of 2
No order, A shakes B same as B shakes A
So {}^8C_2
In how many different ways, can the letters of the word 'INHALE' be arranged ?
INHALE has 6 letters, none repeating.
Total arrangements = 6!
In how many different ways, can the letters of the word 'BANKING' be arranged ?
BANKING has 7 letters, N repeating
Total Arrangements =
\frac{7!}{2!}
In a meeting between two countries, each country has 12 delegates. All the delegates of one country shake hands with all delegates of the other country. Find the number of handshakes possible ?
all x all = 12 x 12
How many can straight lines be drawn from 15 noncollinear points ?
Combination of 2 points. And a line cant be drawn from a point to itself
={}^{15}C_2  15
=105
There are 25 students in a class. Find the number of ways in which a committee of 3 students is to be formed?
Combination as order is not important
{}^{25}C_3
In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?
8I, 4A, 4E
Consider each group as whole and find their seating
3 groups, order important, 3P3 = 3!
In each group, you can have arrangements, order
Total =3! 8! 4! * 4!
How many different words can be formed using all the letters of the word ALLAHABAD?
(a) When vowels occupy the even positions.
(b) Both L do not occur together.
9 letters,
Repeats A 4, L  2,
Vowels = A only, So only one vowel combination
Remaining 5 letters, and arrangements = 5!
Consider both L's together as single letter, then arrangements = 4!
So arrangements where both Ls don't occur together = 5!4!
120  24= 96