Note: The video for this module contains a summary of all the concepts covered in this lesson. The video would serve as a good revision. Please watch this video in intervals of a few weeks so that you do not forget the concepts. Below is a cheatsheet that includes all the formulae but not necessarily the concepts covered in the video.
9. Cheatsheet
1) n!=1×2...×n
2) nPr=(n−r)!n! and nCr=r!(n−r)!n!
4) Permutation of n elements where a elements are of one kind, b of another kind, etc. is a!b!c!...n! ways.
5) Similar to the point above, if in an n-letter word, if 1 letter occurs a times, another letter occurs b times, etc., then number of different n-letter words that can be formed =a!b!c!...n!
6) Using Blanks: Identify the number of elements that can occupy each blank and multiply them.
7) While using blanks, address or fill up the blanks which have conditions first. E.g., in digits-related questions, the first digit from the left will have to be a digit other than zero.
8) As a general rule, multiply when And is used and add when Or is used. (Note: Exceptions exist)
9) If ndistinct items can each be assigned in rdifferent ways, then this can be done in rn ways.
10) Slotting: We arrange elements without conditions first and then select the slots and then arrange the rest.
11) Conditional Permutation: One approach is to first select the items and then arrange. The other approach is to use blanks and apply the conditions.
12) Sum of all n-digit numbers using n distinct non-zero digits =(n−1)!× sum of digits ×(111...ntimes)
13) Path Movement: Use letters to represent the steps or movements and then permute the letters of the word.
14) n distinct elements can be circularly arranged in (n−1)! ways
15) If this circle can be reversed, number of such arrangements is 2(n–1)!
16) In a total of n objects, if r are wrongly assigned,
Number of such derangements =nCr×D(r) = nCr×r!×(1−1!1+2!1−3!1+...)
17) nCr=nCn−r=r!nPr
The greatest value of nCr is r=2n, when n is even and r=2(n+1) or 2(n−1) when, n is odd.
18) nC0+nC1+nC2+...+nCn=2n nC0+nC2+nC4+...=nC1+nC3+nC5+...=2n−1
When n is odd, nC0+nC1+nC2+...+nC(n−1)/2=nC(n+1)/2+nC(n+3)/2+...+nCn=2n−1
19) Selecting Distinct Elements: If n distinct elements are to be divided into groups with a elements, b elements, c elements, etc., number of possible selections is a!b!c!...n!
20) Selecting Identical Elements: From n identical elements, the number of ways to make
- 0 or more selections is n+1 ways; and
- 1 or more selections is n ways.
21) Number of ways nidentical items can be divided into rdistinct groups is n+r−1Cr−1.
22) Number of terms in the expansion of (a1+a2+...+ar)n=n+r−1Cr−1.
Number of terms in the expansion of (1+a+a2+...+ar)n is nr−1.
23) When elements have to be divided into similar groups, then it needs to manually done.
24) Maximum Intersection points of n circles of different radii =n(n−1) and n non-concurrent lines =nC2
25) Maximum lines that can be drawn through 2 or more of non-collinear points =nC2
Number of straight lines drawn where through 2 or more of n points where r are collinear =nC2−rC2+1
26) Number of triangles drawn from n non-collinear points =nC3
Number of triangles drawn from n points, where r points are collinear =nC3−rC3
27) Number of diagonals of an n-sided polygon =2n(n−3)
28) From two sets of m parallel lines and n parallel lines, number of parallelograms formed =mC2×nC2
29) From a larger rectangle with m rows and n columns,
Number of rectangles that can be formed =(1+2+..+m)(1+2+...+n)=2m(m+1)×2n(n+1)
Number of squares that can be formed =mn+(m−1)(n−1)+(m−2)(n−2)+...
30) From a larger square with n rows and n columns,
Number of rectangles that can be formed =(2n(n+1))2
Number of squares that can be formed =n2+(n−1)2+...+1=6n(n+1)(2n+1)
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