Dictionary Rank of a Word | Permutations & Combinations

 PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni


Geometric Progression, Its nth term, Sum to n terms, sum to infinite terms, Properties of G.P.,Special Results
A sequence of non-zero numbers is called a geometric progression (G.P.) if the ratio of the term and the term preceding to it is always a constant quantity.
\[A\:sequence\:a_{1},\: a_{2},\: a_{3},\: a_{4}........,a_{n}, \: a_{n+1}..... is\: called \:geometric\: progression\]\[If \: a_{n+1}=a_{n}\times \: r\: , where \: r \: is\: the\: common \: ratio \: of \: GP \]\[Where\: \: \: r=\frac{a_{n+1}}{a_{n}}\]
General Geometric Progression is\[a,\: ar,\: ar^{2},\: ar^{3},.........ar^{n-1}\]
First term = a, Second term = ar,  and so on and  r is the common ratio of GP
\[nth \: term\: in\: GP\: is\: \: ar^{n-1}\]
Sum of first n terms in GP is \[S_{n}=\frac{a(r^{n}-1)}{r-1}\: where, r > 1\]
Sum of first n terms in GP is\[S_{n}=\frac{a(1-r^{n})}{1-r}\: where,\: \: r < 1\]
Sum to infinite terms in GP.
General GP with infinite number of terms is of the following type
\[a+ar+ar^{2}+ar^{3}+ar^{4}+........+ ar^{n-1}+ .........\]
Sum to infinite terms of GP is given by\[S_{\infty }=\frac{a}{1-r}\; , \; Where -1< r< 1 \; or\;\left | r \right | < 1\] 
\[Three\; terms\; in\; \; G.P.\; are:\; \; \; \frac{a}{r},\; a,\; ar, \; \; here \; common \; ratio\; is\; r\]
\[Four\; terms\; in\; \; G.P.\; are:\; \; \; \frac{a}{r^{3}},\; \frac{a}{r},\; ar, \; ar^{3},\;\; here \; common \; ratio\; is\; r^{2}\]\[Five\; terms\; in\; \; G.P.\; are:\; \; \; \frac{a}{r^{2}},\; \frac{a}{r}, a,\; ar, \; ar^{2},\;\; here \; common \; ratio\; is\; r\]
1) If all the terms of G.P.  be  multiplied  or  divided  by  the  sane  non- zero  constant,    then  it remains a   G.P.   with  common  ratio  r.
2) Reciprocal of the terms of a G.P. form a G.P.
3) If each term of a G.P. raised to the same power , the resulting sequence also form a G.P.
\[4)\; If a_{1},\; a_{2},\; a_{3},........\; a_{n}.... are\; the \; terms \; of \; G.P.,\; \; then\;\\ log\; a_{1},log\; a_{2},.....log\; a_{n}........... \; \;\; \]\[are\; in\; A.P. \; \; and \; \; vice\; -\; versa\]
If a, b, c are the three terms of GP then b is said to be the Geometric Mean and is given by\[b=\sqrt{ac}\: \: or\: \: b^{2}=ac\]
If a, b, c are in GP then\[\frac{b}{a} = r = \frac{c}{b}\] ⇒ \[\frac{b}{a} = \frac{c}{b}\]\[b^{2}=ac\Rightarrow b=\sqrt{ac}\]

Componendo and Dividendo
If four terms a, b, c, d are proportional then \[\frac{a}{b}=\frac{c}{d}\]
When we apply componendo and dividendo then we get  \[\frac{a+b}{a-b}=\frac{c+d}{c-d}\]
\[or\; \; \frac{Numerator+Denomenator}{Numerator-Denomenator} =\frac{Numerator+ Denomenator}{Numerator-Denomenator}\]
Sum of first n  natural number  is given by\[1+2+3+4+5......n=\frac{n(n+1)}{2}\]
Sum of square of first n natural number is given by\[1^{2}+2^{2}+3^{2}+4^{2}+......+ n^{2}=\frac{n(n+1)(2n+1)}{6}\]
Sum of cube of first n natural number is\[1^{3}+2^{3}+3^{3}+4^{3}+......+n^{3}=\left [ \frac{n(n+1)}{2} \right ]^{2}\]

The first term of an infinite G.P. is 1 and any term is equal to the sum of all terms that follow it. Find the infinite G.P.
Solution:  It is given that:   a = 1

A.T.Q.        Tn = Tn+1 + Tn+2 + Tn+3 +  ……………………

                  arn-1 = arn + arn+1 + arn+2 + ……………………

\[ar^{n-1}=\frac{ar^{n}}{1-r}\]Now cross multiply it  and putting a = 1 we get \[r^{n-1}-r^{n}=r^{n}\Rightarrow 2r^{n}=r^{n-1}\]\[\Rightarrow 2r = 1 \Rightarrow r=\frac{1}{2}\]

Putting a = 1 and r = 1/2 we get the required sequence as follows \[\frac{1}{2},\; \frac{1}{4},\; \frac{1}{16},\; .......,\; \infty\]


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