Combinations

We are asked to select 4 numbers from the set $A%20%3D%20%5C%7B1%2C%203%2C%205%2C%207%2C%20%5Cdots%2C%2021%5C%7D$ with the condition that 11 must always be included.
The set $A$ contains 11 odd numbers: $A%20%3D%20%5C%7B%201%2C%203%2C%205%2C%207%2C%209%2C%2011%2C%2013%2C%2015%2C%2017%2C%2019%2C%2021%20%5C%7D$ We are selecting 4 numbers in total, and one of them is fixed as 11.
This leaves us with 3 more numbers to choose from the remaining 10 numbers: $%5C%7B%201%2C%203%2C%205%2C%207%2C%209%2C%2013%2C%2015%2C%2017%2C%2019%2C%2021%20%5C%7D$ The number of ways to select 3 numbers from these 10 is given by the combination formula $%5Cbinom%7B10%7D%7B3%7D$ , which is: $%5Cbinom%7B10%7D%7B3%7D%20%3D%20%5Cfrac%7B10%20%5Ctimes%209%20%5Ctimes%208%7D%7B3%20%5Ctimes%202%20%5Ctimes%201%7D%20%3D%20120$ Thus, the correct answer is (A).

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Combinations

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs