Problems in Combinations

Here are more explainations about Combinations. Problems and solutions for you :)

Definition

" The different selections possible from A collection of items are called combinations " 

For example:

The different selections possible from the alphabets A, B, C, taken 2 at a time, are AB, BC and CA.
It does not matter whether we select A after B or B after A. The order of selection is not important in combinations.

To find the number of combinations possible from a given group of items n, taken r at a time, the formula, denoted by:

n C r is
n C r = n! / [r! * (n-r)!]

For example, verifying the above example, the different selections possible from the alphabets A, B, C, taken two at a time are:

3 C2 = 3! / (2! * (3-2)!) = 3 possible selections (i.e., AB, BC, CA)

Important Combination formula:

n C n = 1
n C 0 = 1
n C 1 = n
n C r = n C(n-r)

The number of selections possible with A, B, C, taken all at a time is 3C 3 = 1 (i.e. ABC)

Solved examples of Combination.

Let us take a look at some examples to understand how Combinations work:

Problem 1: In how many ways can a committee of 1 man and 3 women can be formed from a group of 3 men and 4 women?

Solution :

No. of ways 1 man can be selected from a group of 3 men =
3 C1 = 3! / 1!*(3-1)! = 3 ways.
No. of ways 3 women can be selected from a group of 4 women = 4C 3 = 4! / (3!*1!) = 4 ways.
Problem 2: Among a set of 5 black balls and 3 red balls, how many selections of 5 balls can be made such that at least 3 of them are black balls.

Solution :

Selecting at least 3 black balls from a set of 5 black balls in a total selection of 5 balls can be:

3 B and 2 R
4 B and 1 R and
5 B and 0 R balls.

Therefore, our solution expression looks like this.
5 C3 * 3C 2 + 5 C4 * 3C 1 + 5 C5 * 3C 0 = 46 ways .
Problem 3: How many 4 digit numbers that are divisible by 10 can be formed from the numbers 3, 5, 7, 8, 9, 0 such that no number repeats?

Solution :

If a number is divisible by 10, its units place should contain a 0.

_ _ _ 0

After 0 is placed in the units place, the tens place can be filled with any of the other 5 digits.

Selecting one digit out of 5 digits can be done in 5C 1 = 5 ways.

After filling the tens place, we are left with 4 digits. Selecting 1 digit out of 4 digits can be done in 4C 1 = 4 ways.

After filling the hundreds place, the thousands place can be filled in 3C 1 = 3 ways.

Therefore, the total combinations possible = 5*4*3 = 60.