Combination methods related math calculation are given in bank exams so its important to learn for exams We can’t ignore it. So it is very very important to you improve your maths skills for banking exams in combination methods.
Shortcut tricks on combination are one of the most important topics in exams. Time is the main factor in competitive exams. If you manage your time then you can do well in those exams. Most of us skip that part. Here in this page we give few examples on Combination shortcut tricks. These shortcut tricks cover all sorts of tricks on Combination. Visitors are requested to carefully read all shortcut examples. You can understand shortcut tricks on Combination by these examples.
Before doing anything we recommend you to do a math practice set. Write down twenty math problems related to this topic on a page. Using basic math formula do first ten maths of that page. You also need to keep track of timing. After solving all ten math questions write down total time taken by you to solve those questions. Now go through our page for combination shortcut trick. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of Timing. You will surely see the improvement in your timing this time. But this is not all you want. You need more practice to improve your timing more.
We all know that the most important thing in competitive exams is Mathematics. That doesn’t mean that other sections are not so important. But if you need a good score in exam then you have to score good in maths. Only practice and practice can give you a good score. You should do your math problems within time with correctness, and you can do this only by using shortcut tricks. Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential that you may do maths within time without using any shortcut tricks.
But, other peoples may not do the same. For those we prepared this combination shortcut tricks. We always try to put all shortcut methods of the given topic. But if you see any tricks are missing from the list then please inform us. Your little help will help so many needy.
In short we say combination is a group, selection and Each of the different groups or selections which can be made by taking some number of objects or all of a number of objects, that is called a combination. In combination we can select only one item at time it based on selection on choice. Suppose Ball, Marbel, etc.
Types of Combination?
Type #1
Lets we want to select two women out of three women A, B, C. Then possible selection would be AB, BC and CA.
Note: Here AB and BA represents the same selection.
Type #2
Suppose we wants to select all three girls between A, B, C. Then the selection would be ABC.
Note: ABC, ACB, BAC, BCA, CAB, CBA all are the same choice of selection from that.
Type #3
Various person of two out of four person A, B, C, D are : AB, AC, AD, BC, BD, CD.
The number of all combinations of n things, taken r at a time is:
n C r = n! / r! ( n – r ) ! = n ( n – 1 ) ( n – 2 ) …….to r factors / r!.
Note: if n = r , nCr = 1 and n C 0 = 1.
nCr = n C (n – r ). = 1.
Ex: 16 C 13 = 16 C ( 16 – 13 ) = 16 C 3 = 16 x 15 x 14 x 13! / 3! = 16 x 15 x 14 / 3 x 2 x 1 = 560
Find out value of 10 P 3
= 10 C 3= 10 x 9 x 8 / 3! = 10 x 9 x 8 / 3 x 2 x 1 = 120
Find the value of 50 C 50
= 50 C 50 = 1. [ n C n = 1]
In how many different ways can 7 player selected from a group of 10?
Show Answer Show How to Solve Open Rough Workspace
In how many different ways can a football team of 11 players is selected out of 16 players, and if one player is to be excluded?
Show Answer Show How to Solve Open Rough Workspace
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thank you for giving me very good suggestions!
I love this website!
taught me really well!
made me chapter easy!
🙂
anybdy can tell me how can i got velues of ‘r’
whene n is avalible
with any example
for eg; 5C2=10. What if the problem said nC4=10 , find the value of n ?
here is the basic concept :
5C2=10
nC2 =10
we know that , nCr = n!/r!(n-r)!
likely,
nC2 =10
n!/2!*(n-2)! = 10
n!/(n-2)! = 10 * 2!
n!/(n-2)! = 20 { because 2! = 1*2 }
to cancel (n-2)! we can change the dinominator in to
ie; n! = n*(n-1)*(n-2)!
there for both (n-2)! get cancelled and it becomes
n*(n-1) = 20
opening the bracket of l.h.s we get
n^2 – n = 20
ie; n^2-n-20 = 0
n^2 – 5n+4n – 20 = 0
(n-5) (n+4) = 0
ie; n=5 or n= -4
either n can be -4 or 5 , since we are calculating for combionations there can be no negative values , hence the answer is 5.
baji ka number dey phir hi r ki value khudi ajae gi
Really good…gonna help a lot to the students..
I’ve got a question and it asks me to find how many numbers between 2500 and 5000 can be formed using digits 1,2,3,4,5 and 7 with no digit being repeated, how do i do this?
The answer would be 144.
the number must be between 2500 to 5000. so
lets start with
fixing 25_ _
we have 4 digits to use so = 4*3=12.
now here we don’t have 6 number so we cannot form number with 26_ _.
next would be 27_ _ = 4 * 3 = 12.
now we can have from 3 _ _ _ . we again have 5 numbers to use . Thus = 5 * 4 * 3 = 60.
for numbers statring with 4 _ _ _. again same = 5 * 4 *3 = 60.
Thus total combinations possible would be = 12+12+60+60 = 144.
Can you please explain combination example 1 once again, because i didn’t get how this (10,7) possible for required no of ways..
Use the formula “n!/r!( n-r)!” .
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