vyomb wrote:a single person can have the following options
amount of prize received=0 or 1 or 2 or 3 or 4
so total 5 possibilities.

hence for 4 people we have 4 pow 5=1024 ways.

Vinod R wrote:Hi,
Solutions for 3 questions among the given set of questuions are as follows:
1. If 2nC3: nC2 = 44:3 find n.

Sol: Expand the two 2nC3 and nC2 with standard formulae:
2n!/3!*(2n-3)! /n!/2!*(n-2)! = 44/3
2n(2n-1)(2n-2)(2n-3)!/3!*(2n-3)!/n*(n-1)*(n-2)!/2!*(n-2)! = 44/3
Now (2n-3)! and (n-2)! will be removed.

2n(2n-1)(2n-2)*2!/n*(n-1)*3! = 44/3
Solving the above u will get
2n-1 = 11
n =6.

2. If nC12 = nC8 what is the value of 22Cn. (Ans: 231)

Sol: Solution to this problem is quite simple as it is based on the standard formulae :
nCr = nCn-r

Just make a simple guess :
For satisfying the above standard formulae "20" will be the right choice. So substitute 20 then

20C12 = 20C8
So 22C20 = 231

3. If 4 times the number of permutations of n things 3 together is equal to 5 times the number of permutations of n-1 things 3 together find n.

Sol: The above question can be written as:

4(nP3)= 5.((n-1)P3)

By solving the above u wil get:
4n = 5n - 15;
n = 15.

Regards,
Vinod.

fireplayer wrote:i am confused with the last question, how can be the ans= 1024 when everyone eligible to get one or more than one prize? someone will get none if other get 3 prize?