![]() So 252 different 5 person committees can be elected from 10 people. 8) How many different five-person committees can be elected from 10 people? In 4 different combination you get any three of them. Solution: here n=4, r=3 this is 4 CHOOSE 3 problem ![]() 7) Find the number of combinations for 4 choose 3. In 20 different combination you get any three of them. How many unlike ways can you use to draw only three colour? 6) Box of crayons having red, blue, yellow, orange, green and purple. In 10 different combination you get any three of them. Solution: here n=5, r=3 this is 5 CHOOSE 3 problem 5) Find the number of ways of choosing 3 members from 5 different partners. In 15 different combination you get any two of them. Solution: here n=6, r=2 this is 6 CHOOSE 2 problem 4) Find the number of combinations 6 choose 2. In 10 different combination you get any two of them. Solution: here n=5, r=2 this is 5 CHOOSE 2 problem ![]() ![]() 3) How many different combinations can be made if you have only 5 characters and you have to choose any 2 among them? In 6 different combination you get any two of them. Solution: here n=4, r=2 this is 4 CHOOSE 2 problem 2) How many different combinations can be done if you have 4 different items and you have to choose 2? In 3 different combination you get any two of them. Solution: here n=3, r=2 this is 3 CHOOSE 2 problem NC k = n!/k!(n-k)! Examples: 1) If you have three dress with different colour in red, yellow and white then can you find a different combination you get if you have to choose any two of them? This is commonly known as nCr or n choose k formula. The number of combinations (selections or groups) that can be set up from n different objects taken r (0<=r<=n) at a time is Examples based on Combination (nCr formula/ n choose k formula) In 5040 number of ways, 7 people can organize themselves in a row. Example 6) Find the number of ways, 7 people can organize themselves in a row. Example 5) Find out the number of ways a judge can award a first, second, and third place in a contest with 18 competitors.ġ8 P 3 = 18!/(18-3)! = 15!.16.17.18/15! = 4896Īmong the 18 contestants, in 4896 number of ways, a judge can award a 1st, 2nd and 3rd place in a contest. In 720 ways, three-digit permutations are available. Example 4) How many different three-digit permutations are available, selected from ten digits from 0 to 9 combined?(including 0 and 9). In 358800 ways, 4 different letter permutations are available. Example 3) How many permutations are possible from 4 different letter, selected from the twenty-six letters of the alphabet?Ģ6 P 4 = 26!/(26-4)! = 22!.23.24.25.26/22! = 358800 In 5040 ways 4 women can be chosen as team leaders. NP r= n!/( n-r)! Example 1): There is a train whose 7 seats are kept empty, then how many ways can three passengers sit. The number of ways of arranging = The number of ways of filling r places. Here We are making group of n different objects, selected r at a time equivalent to filling r places from n things. The methods of arranging or selecting a small or equal number of people or items at a time from a group of people or items provided with due consideration to be arranged in order of planning or selection are called permutations.Įach different group or selection that can be created by taking some or all of the items, no matter how they are organized, is called a combination. In this article, we have discussed some examples which will make the foundation strong of the students on Permutations and Combinations to get the insight clearance of the concept, it is well aware that the Permutations and combinations both are the process to calculate the possibilities, the difference between them is whether order matters or not, so here by going through the number of examples we will get clear the confusion where to use which one. Illustration of the concept Permutations and Combinations by the examples ![]()
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