GMAT 2020 NEWS
Test takers take GMAT for securing admissions in the top business schools of the world. Quant section of GMAT has always topped the headline because of its tricky nature and that is quite evident from the arithmetic section. Permutation and combination are one of the topics in GMAT’s quant section and there are certain people who have difficulty solving it. Definitely to mentioned about the time crunch that keeps people away from reaching the right answer.
But few preparation tips will cover all the major problems of permutation and combination and will help you get a good GMAT score. Permutation and combination is the way of finding out the number of myriad outcomes placed in a given set of items and numbers of limitations. It all depends on whether the syncing matters. If you are still confused (just like we were initially), let us explain an example to ease it out: imagine having three chips packets of different flavors – a tomato flavored, a chilli flavored and a cheese flavored. If you are asked to pick two of them, how many possibilities are left to you? If you are thinking three, then congrats! You are already on the right track. The combination of tomato with butter, butter with chilli and chilli with tomato – and that is exactly what combination is. On the other hand, if you are asked about the manner of eating or priority-based eating then the options will be even more. You can have the tomato first and then the chilli, for instance – and that is what permutation is. Let us now discuss a few of the important sections of this topic:
Let us first start with the adding part or concept 1 of permutation and combination – the importance of the word ‘OR’ has to be studied here. Let us take two situations for these –
Lucy has 3 types of jeans and 2 different kinds of shirts. Whenever she goes out she prefers wearing jeans with a particular shirt – in how many manners can she wear it?
In the second instance, Peter has 3 sneakers and 2 flip flops. Whenever he hangs out with his friends he likes wearing either the sneaker or any of the flip flop - in how many ways can he wear it?
If you look carefully, in both the scenarios, the numbers considered are similar. The only distinguishing factor is – while Peter will be able to put on a single pair of either a sneaker or a flip-flop, Lucy can wear two pieces, namely, jeans along with a shirt.
Further explanation of Peter’s scene: He has three differently colored sneakers: blue, black, white and two differently colored flip-flops, red and green. When he has to choose something, he has to go for -
Either the blue sneaker OR,
The black sneaker OR,
The white sneaker OR,
The green flip-flop OR,
The red flip-flop.
The above explanation solves one question that Peter cannot put on a sneaker and a flip-flop at the same time – so he doesn’t have the option to wear two different things. The answer will, therefore, be 5.
By the above explanation, we can further proceed with the fact that whenever we have to answer this kind of question, we’ve to carefully read whether it is an ‘OR’ or ‘AND’ mentioned. If the question is – Peter has three differently colored sneakers and two differently colored flip-flops – he prefers wearing either a sneaker OR a flip-flop while going out, in how many ways can he wear that?
As we can see the word OR in the question, we can ascertain that both the events cannot be clubbed together. Whenever there is an instance where both the events cannot be clubbed together, we simply add them. Therefore to solve this:
In total possible cases for Peter – he will either wear sneakers or flip-flops,
Number of manners he can wear sneakers: 3,
Number of manners he can wear flip-flops: 2,
Therefore, total possible cases: 3 OR 2 = 3+2 = 5.
Let us proceed with the multiplying part or the usage of the keyword ‘AND’.
Now comes Lucy who has three different jeans: a blue one, a black one and a grey one; and a red shirt and a white shirt. Let us list down the possible ways she can pair them up:
Now let us consider the question: Lucy has three different jeans and two different shirts, whenever she steps out she prefers wearing a jeans AND a shirt, in how many possible ways can she pair them up?
Notice the use of the word AND. It can thus be understood that in no circumstances cans she wear either a shirt OR jeans but always have to wear them together. Whenever we have situations like these, we always multiply.
Total possible cases for Lucy: she can wear jeans AND a shirt,
Number of ways she can wear jeans (blue OR black OR grey) [notice the word OR which means addition];
Number of ways she can wear a shirt (Red or White) [notice OR which means addition];
Therefore, 3 AND 2 = 3 * 2 = 6.
Dealing with Listing Problems (h3)
To understand the listing problem, let us take: a, b, c and d and you’re supposed to pick two out of them.
Let us pick the first item a and combine it with b. gradually climbing down to two more items, namely, c and d and similarly combining with them: ab, ac, ad.
Similarly, combine b with the remaining but not with a. So coming down to bc and bd.
In the same manner, do it with c, but not with a or b. so coming down to cd.
You have to keep moving on till you reach the second last item and finally, you have to count if you have received these items: ab, ac, bc, bd, cd.
Finally, it’s an end to the listing problem.
After listing the next thing that comes is a tree diagram. The tree diagram is the diagrammatic method of solving which is similar to the listing one. Now let us use the previous 4 items as for example. The last line of the tree diagram will show 12 boxes which will display possible ways of permuting 2 items out of 4. Each item can be clubbed with 3 other things. The other way of solving this is – imagine you have two blank spaces to be filled _ and _, in which the first space can be comprised of 4 possible items and the second space with 3 possible items. As you’ve already placed one in the first space so it is 4 * 3 = 12.
Quick-Fix to Fundamental Counting Principle
After the tree diagram, let us study the fundamental counting principle which is indeed an interesting one. This method states: if task A can be done in ‘m’ number of ways and after completion if task ‘b’ can be done in ‘n’ number of ways – this concludes in m*n number of ways to complete both tasks A and B.
Let us consider the word: M A N G O
If you are thinking the answer to be 120 then congrats, you’re already on the right path.
Let us break it down: Imagine 5 spaces _ _ _ _ _
In the first blank space, you can put either of 5 alphabets, in the second space leaving that first one you can put 4 more and accordingly, in the third space, you can exclude those previously placed two items and so on.
Lastly, the answer will be 5*4*3*2*1 or 120!
Putting the formulas:
So now, when we are aware of the main ways of finding permutation and combinations or if we have to set it according to the order or choose – now it is time to put them in the formulas.
The various ways of permuting ‘r’ object by ‘n’ object is provided by:
n P r = n!/(n-r)! where
n! = n * (n-1) * (n-2) ….. * (2) * (1)
Note: 0! is defined as 1.
How many paths are there of arranging 4 letters from the word O R A N G E?
The word ORANGE has 6 different alphabets and thus the order matters and so we have to permute. When the formula is applied:
6 P 4 = 6!/(6-4)! – 6!/2! – 360.
Things to remember:
The time factor might scare you while solving the question but never fret. This will only make you lose marks the answer to the problem which you already knew.
Many are unaccustomed to the computer-adaptive test and one needs to practice mock tests before sitting for the final exam, therefore.
Always remembering the fact that test makers will try to confuse the students and that is inevitable but what can be done is to keep the fear away and having confidence (not overconfidence though).
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