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Permutation and CombinationQuestion and Answers: Page 20

Question Number 54740 Answers: 3 Comments: 0

Prove that 1. ((n),(r) ) = (((n−1)),(( r)) ) + (((n−1)),(( r−1)) ) 2. ((n),(r) ) + ((( n)),((r−1)) ) = (((n+1)),(( r)) ) 3. ((n),(0) )+ ((n),(1) )+ ((n),(2) )+..+ ((n),(n) )=2^n

$${Prove}\:{that} \\ $$$$\mathrm{1}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\:+\:\begin{pmatrix}{{n}−\mathrm{1}}\\{\:{r}−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{2}.\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix}\:+\begin{pmatrix}{\:\:\:{n}}\\{{r}−\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\:{r}}\end{pmatrix}\: \\ $$$$\mathrm{3}.\:\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}+..+\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}=\mathrm{2}^{{n}} \\ $$

Question Number 54698 Answers: 2 Comments: 0

How many words with at least 2 letters can be formed using the letters from TINKUTARA?

$${How}\:{many}\:{words}\:{with}\:{at}\:{least}\:\mathrm{2}\:{letters} \\ $$$${can}\:{be}\:{formed}\:{using}\:{the}\:{letters}\:{from} \\ $$$${TINKUTARA}? \\ $$

Question Number 54646 Answers: 2 Comments: 0

Such That a. _(n+1) C_r =(((n+1). _n C_r )/((n−r+1))) b. _n C_0 +_n C_2 +_n C_(4...) =_n C_1 +_n C_3 +_n C_(5...) =2^(n−1)

$$\mathrm{Such}\:\mathrm{That} \\ $$$$\mathrm{a}.\:_{{n}+\mathrm{1}} {C}_{{r}} =\frac{\left({n}+\mathrm{1}\right).\:_{{n}} {C}_{{r}} }{\left({n}−{r}+\mathrm{1}\right)} \\ $$$$\mathrm{b}.\:_{{n}} {C}_{\mathrm{0}} +_{{n}} {C}_{\mathrm{2}} +_{{n}} {C}_{\mathrm{4}...} =_{{n}} {C}_{\mathrm{1}} +_{{n}} {C}_{\mathrm{3}} +_{{n}} {C}_{\mathrm{5}...} =\mathrm{2}^{{n}−\mathrm{1}} \\ $$$$ \\ $$

Question Number 54286 Answers: 2 Comments: 2

Question Number 53043 Answers: 2 Comments: 1

Question Number 52381 Answers: 2 Comments: 0

If all the words with or without meaning are written using the letters of the word QUEEN and are arranged as in English dictionary.What is the position of the word QUEEN? a)44th b)45th c)46th d)47th

$${If}\:{all}\:{the}\:{words}\:{with}\:{or}\:{without} \\ $$$${meaning}\:{are}\:{written}\:{using}\:{the} \\ $$$${letters}\:{of}\:{the}\:{word}\:{QUEEN}\:{and} \\ $$$${are}\:{arranged}\:{as}\:{in}\:{English} \\ $$$${dictionary}.{What}\:{is}\:{the}\:{position} \\ $$$${of}\:{the}\:{word}\:{QUEEN}? \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\mathrm{4}{th}\:{b}\right)\mathrm{45}{th}\:{c}\right)\mathrm{46}{th}\:{d}\right)\mathrm{47}{th} \\ $$

Question Number 52379 Answers: 1 Comments: 0

The sum of the digits in the unit place of all four digit numbers formed using the numbers 3 4 5 andL 6 without repetition is a)432 b)108 c)36 d)18

$${The}\:{sum}\:{of}\:{the}\:{digits}\:{in}\:{the}\:{unit} \\ $$$${place}\:{of}\:{all}\:{four}\:{digit}\:{numbers} \\ $$$${formed}\:{using}\:{the}\:{numbers}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:{and}\mathscr{L} \\ $$$$\mathrm{6}\:{without}\:{repetition}\:{is} \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\mathrm{32}\:{b}\right)\mathrm{108}\:{c}\right)\mathrm{36}\:{d}\right)\mathrm{18} \\ $$

Question Number 52278 Answers: 2 Comments: 0

8 digit numbers are formed using 1 1 2 2 2 3 4 4. The number of such numbers in which the odd digits do not occupy odd places is a)160 b)120 c)60 d)48

$$\mathrm{8}\:{digit}\:{numbers}\:{are}\:{formed}\:{using} \\ $$$$\mathrm{1}\:\mathrm{1}\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}.\:{The}\:{number}\:{of}\:{such} \\ $$$${numbers}\:{in}\:{which}\:{the}\:{odd}\:{digits} \\ $$$${do}\:{not}\:{occupy}\:{odd}\:{places}\:{is} \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\mathrm{60}\:{b}\right)\mathrm{120}\:{c}\right)\mathrm{60}\:{d}\right)\mathrm{48} \\ $$

Question Number 52276 Answers: 1 Comments: 0

An 8 digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is? a)72(7i) b)18(7i) c)40(7i) d)36(7i) please I mean factorial by i.The part of my screen where the factorial symbol is isnt functioning.Thanks

$${An}\:\mathrm{8}\:{digit}\:{number}\:{divisible}\:{by}\:\mathrm{9}\:{is} \\ $$$${to}\:{be}\:{formed}\:{using}\:{digits}\:{from} \\ $$$$\mathrm{0}\:{to}\:\mathrm{9}\:{without}\:{repeating}\:{the}\:{digits}. \\ $$$${The}\:{number}\:{of}\:{ways}\:{in}\:{which}\:{this} \\ $$$${can}\:{be}\:{done}\:{is}? \\ $$$$\left.{a}\left.\right)\left.\mathrm{7}\left.\mathrm{2}\left(\mathrm{7}{i}\right)\:{b}\right)\mathrm{18}\left(\mathrm{7}{i}\right)\:{c}\right)\mathrm{40}\left(\mathrm{7}{i}\right)\:{d}\right)\mathrm{36}\left(\mathrm{7}{i}\right) \\ $$$$ \\ $$$${please}\:{I}\:{mean}\:\boldsymbol{{factorial}}\:{by}\:{i}.{The} \\ $$$${part}\:{of}\:{my}\:{screen}\:{where}\:{the}\:{factorial} \\ $$$${symbol}\:{is}\:{isnt}\:{functioning}.{Thanks} \\ $$

Question Number 51779 Answers: 0 Comments: 3

Q= In how many ways 8 students can be divided into two equal groups? Ans= ((8!)/(2!×(4!)^2 )) here why two factorial is divided? we do not do the same incase of distributing 52 cards equally among 4 persons. pls explain...

$${Q}=\:\:{In}\:{how}\:{many}\:{ways}\:\mathrm{8}\:{students}\:{can}\:{be}\:{divided}\:{into}\:{two}\:{equal}\:{groups}? \\ $$$$ \\ $$$${Ans}=\:\:\:\:\frac{\mathrm{8}!}{\mathrm{2}!×\left(\mathrm{4}!\right)^{\mathrm{2}} }\:\:\: \\ $$$$ \\ $$$${here}\:{why}\:{two}\:{factorial}\:{is}\:{divided}? \\ $$$${we}\:{do}\:{not}\:{do}\:{the}\:{same}\:{incase}\:{of}\:{distributing}\:\mathrm{52}\:{cards}\:{equally}\:{among}\:\mathrm{4}\:{persons}. \\ $$$$ \\ $$$${pls}\:{explain}... \\ $$

Question Number 51465 Answers: 2 Comments: 4

Question Number 51287 Answers: 1 Comments: 0

Question Number 51286 Answers: 1 Comments: 0

Question Number 50724 Answers: 1 Comments: 0

a man goes in for an examination in which there are 4 papers which maxmum of 10 marks for each paper the no of ways of getting 20 marks on the whole is ans:891

$$\mathrm{a}\:\mathrm{man}\:\mathrm{goes}\:\mathrm{in}\:\mathrm{for}\:\mathrm{an}\:\mathrm{examination}\:\mathrm{in} \\ $$$$\mathrm{which}\:\mathrm{there}\:\mathrm{are}\:\mathrm{4}\:\mathrm{papers}\:\mathrm{which}\:\mathrm{maxmum} \\ $$$$\mathrm{of}\:\mathrm{10}\:\mathrm{marks}\:\mathrm{for}\:\mathrm{each}\:\mathrm{paper}\:\mathrm{the}\:\mathrm{no}\:\mathrm{of}\:\mathrm{ways} \\ $$$$\mathrm{of}\:\mathrm{getting}\:\mathrm{20}\:\mathrm{marks}\:\mathrm{on}\:\mathrm{the}\:\mathrm{whole}\:\mathrm{is} \\ $$$$\mathrm{ans}:\mathrm{891} \\ $$

Question Number 50258 Answers: 1 Comments: 0

A delegation of 4 students is to be selected from a total of 12 students. In how many ways can the delegation be selected if: 1) 2 particular students wish to be included together only in the delegation? 2) 2 particular students refuse to be together and 2 other particular students wish to be together only in the delegation?

$${A}\:{delegation}\:{of}\:\mathrm{4}\:{students}\:{is}\:{to}\:{be} \\ $$$${selected}\:{from}\:{a}\:{total}\:{of}\:\mathrm{12}\:{students}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{the}\:{delegation} \\ $$$${be}\:{selected}\:{if}: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{2}\:{particular}\:{students}\:{wish}\:{to}\:{be}\: \\ $$$${included}\:{together}\:{only}\:{in}\:{the}\:{delegation}? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{2}\:{particular}\:{students}\:{refuse}\:{to}\:{be}\: \\ $$$${together}\:{and}\:\mathrm{2}\:{other}\:{particular}\:{students} \\ $$$${wish}\:{to}\:{be}\:{together}\:{only}\:{in}\:{the}\:{delegation}? \\ $$

Question Number 50008 Answers: 1 Comments: 4

5−digit number divisible by 3 formed using 0,1,2,3,4,5 without repetition. Total number of such no.s is ?

$$\mathrm{5}−{digit}\:{number}\:{divisible}\:{by}\:\mathrm{3}\:{formed} \\ $$$${using}\:\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\:{without}\:{repetition}. \\ $$$${Total}\:{number}\:{of}\:{such}\:{no}.{s}\:{is}\:? \\ $$

Question Number 49982 Answers: 1 Comments: 0

If (n_P_r /n_C_r ) −1=5, find the value of r

$$\mathrm{If}\:\frac{\mathrm{n}_{\mathrm{P}_{\mathrm{r}} } }{\mathrm{n}_{\mathrm{C}_{\mathrm{r}} } }\:−\mathrm{1}=\mathrm{5},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{r} \\ $$

Question Number 49482 Answers: 1 Comments: 14

Find number of 4−letter words which can be formed using the letters of the word ′ALLAHABAD′ such that: NO repetition of letter and word should start “either” from H “or” ends with D ?

$${Find}\:{number}\:{of}\:\mathrm{4}−{letter}\:{words}\:{which} \\ $$$${can}\:{be}\:{formed}\:{using}\:{the}\:{letters}\:{of}\: \\ $$$${the}\:{word}\:'{ALLAHABAD}'\:{such}\:{that}: \\ $$$${NO}\:{repetition}\:{of}\:{letter}\:{and}\:{word}\:{should} \\ $$$${start}\:``{either}''\:{from}\:{H}\:``{or}''\:{ends}\:{with}\:{D}\:? \\ $$

Question Number 49408 Answers: 1 Comments: 3

Find the number of all such 7−digit no. which satisfy conditions: 1) divisible by 3 2) repetition allowed 3) zero is not used.

$${Find}\:{the}\:{number}\:{of}\:{all}\:{such}\:\mathrm{7}−{digit} \\ $$$${no}.\:{which}\:{satisfy}\:{conditions}: \\ $$$$\left.\mathrm{1}\right)\:{divisible}\:{by}\:\mathrm{3} \\ $$$$\left.\mathrm{2}\right)\:{repetition}\:{allowed} \\ $$$$\left.\mathrm{3}\right)\:{zero}\:{is}\:{not}\:{used}. \\ $$

Question Number 49101 Answers: 0 Comments: 0