![]() If the answer to any of the above questions is yes, then in this e-GMAT article you will learn how to get rid of the confusion by using “AND” & “OR” on these questions.ġ.ĝive into the details of the application of “AND” & “OR” in permutation and combination.Ģ.ĝiscuss the attributes that AND – OR present through a few GMAT like questionsģ. ĝo you often get questions incorrect simply because you added entities instead of multiplying them?.Ěre you someone who is confused whether to add or multiply the cases while solving permutation and combination questions?.Trickier problems will involve more than just an outright calculation like those above. Now, if you're dropping them all into one box and you only care about which balls we chose, then t hen we divide by 4!, which is the number of ways to arrange 4 things. Suppose we have 9 distinct balls and you are choosing 4 to drop into 4 distinct boxes. So we just need to take the permutations calculation and divide by r!.įor 8C3, then, take 8 * 7 * 6 and divide by 3!. Since we want to count these different arrangements of r things as the same, simply using a permutations calculation gives us r! too many things. NCr = n!/((n-r)!r!) <- notice this is the same as nPr except with an extra r! at the bottom! This makes sense, because there are r! ways to arrange the r things we choose. In the same way, the combinations formula is the same thing as starting with the FCP and then dividing by the number of positions you have. When we evaluate this, 8! = 8 * 7 * 6 * (5 * 4 * 3 * 2 * 1) and the (5 * 4 * 3 * 2 * 1) cancels out with the 5! in the denominator. That's why there is no need to ever use the permutations "formula:"įor example, 8P3 above is equivalent to 8 * 7 * 6. Now, remember, the permutation formula is equivalent to multiplying the choices for each stage using the Fundamental Counting Principle. If the order/position/role of the things we are choosing are distinct, then we have a permutation. If we only care about what things we choose, then we only care about the combination. To go back to the "people in a room example," we now no longer care only about who is in the room. We care about how they are arranged in the room, or in what order they went into the room. Therefore, in (2), both which 3 people we choose and also the order in which we arrange them matters, which means that (1) is a Permutation/FCP problem.Įquivalent to assigning different positions could be assigning the people to different seats, or cities, etc. 8 choices for President, then 7 choices for VP, then 6 choices for Treasurer. It's 8P3 = 8!/5! This is equivalent to using the Fundamental Counting Principle: 8 * 7 * 6. ![]() So our answer to (2) is 3! times the answer to (1). Now, even with the same combination of 3 people, there are 3! ways to assign these people the different positions:Īnn = Pres, Bob = VP, and Cal = Treasurer is DIFFERENT thanĬal = Pres, Ann = VP, and Bob = Treasurer. We only care who is in the room.įor that reason, (1) is a combination problem.īUT IN PROBLEM (2) we are not only choosing 3 people, but we are giving them 3 different positions: ![]() We don't care about their roles, their seating, or what they are doing. There are actually 3! = 6 different ordering of these three people.īut for (1), all these orderings don't matter - they all represent the same ONE combination of Ann, Bob, and Cal.Īnother way to think about it: We choose three people and put them together in a room. Choosing Ann, Bob, Cal is the same as choosing Bob, Cal, Ann, or Cal, Ann, Bob, etc. In problem (1), there is only one possible combination with each group of 3 people. In both of these problems, we are choosing 3 people out of a group of 8 people. Think about the difference between two problems:ġ) How many ways can we make a team of 3 people from 8 people?Ģ) How many ways can we make a team consisting of a President, VP, and Treasurer from 8 people? ![]() What's the difference between a combination and a permutation? Let's contrast two problems:
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