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Permutation and CombinationQuestion and Answers: Page 7

Question Number 151823 Answers: 2 Comments: 0

Question Number 151615 Answers: 1 Comments: 0

How many numbers greater than 200 can be formed from the digits 1,2,3,4,5 if no digit is to be repeated in any particular number?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{200}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\:\mathrm{if}\:\mathrm{no} \\ $$$$\mathrm{digit}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{repeated}\:\mathrm{in}\:\mathrm{any}\:\mathrm{particular} \\ $$$$\mathrm{number}? \\ $$$$ \\ $$

Question Number 151485 Answers: 1 Comments: 0

Question Number 150859 Answers: 0 Comments: 0

Question Number 150554 Answers: 1 Comments: 0

Show that n_C_r =((n(n−1)(n−2)...(n−r+1))/(r!))

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{n}_{\mathrm{C}_{\mathrm{r}} } =\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)...\left(\mathrm{n}−\mathrm{r}+\mathrm{1}\right)}{\mathrm{r}!} \\ $$

Question Number 149955 Answers: 1 Comments: 0

If a group consist of 8 men and 6 women, in how many ways can a committee of 5 be selected if: i) the committee is to consist of 3 men and 3 women. ii) there are no restrictions on the number of men and women on the committee. iii) at least one man

$$\mathrm{If}\:\mathrm{a}\:\mathrm{group}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{8}\:\mathrm{men}\:\mathrm{and}\:\mathrm{6}\:\mathrm{women}, \\ $$$$\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{a}\:\mathrm{committee}\:\mathrm{of} \\ $$$$\mathrm{5}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{if}: \\ $$$$\left.\:\:\:\:\:\:\mathrm{i}\right)\:\mathrm{the}\:\mathrm{committee}\:\mathrm{is}\:\mathrm{to}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{3}\:\mathrm{men} \\ $$$$\mathrm{and}\:\mathrm{3}\:\mathrm{women}. \\ $$$$\left.\:\:\:\:\:\:\mathrm{ii}\right)\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{restrictions}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{men}\:\mathrm{and}\:\mathrm{women}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{committee}. \\ $$$$\left.\:\:\:\:\:\:\:\mathrm{iii}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{man} \\ $$

Question Number 149595 Answers: 2 Comments: 1

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f (x ):= 3x +[ (x/4) ] , D_( f) =[ 0,∞) f^( −1) ( x )= ?

$$ \\ $$$$\:\:{f}\:\left({x}\:\right):=\:\mathrm{3}{x}\:+\left[\:\frac{{x}}{\mathrm{4}}\:\right]\:\:,\:\mathrm{D}_{\:{f}} \:=\left[\:\mathrm{0},\infty\right) \\ $$$$\:\:\:\:\:\:\:\:\:{f}^{\:−\mathrm{1}} \left(\:{x}\:\right)=\:? \\ $$$$ \\ $$

Question Number 147876 Answers: 1 Comments: 0

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Question Number 145104 Answers: 1 Comments: 0

If ((!6)/x) −!4 = !x then x =?

$$\:\mathrm{If}\:\frac{!\mathrm{6}}{\mathrm{x}}\:−!\mathrm{4}\:=\:!\mathrm{x}\:\mathrm{then}\:\mathrm{x}\:=? \\ $$

Question Number 143740 Answers: 1 Comments: 0

Prove that lim_(n→+∞) 2n−(2n+1)ln(n)+Σ_(p=0) ^n ln(1+p^2 )= ln(e^π −e^(−π) )

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}2n}−\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{n}\right)+\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)=\:\mathrm{ln}\left({e}^{\pi} −{e}^{−\pi} \right) \\ $$

Question Number 143627 Answers: 2 Comments: 0

Question Number 143462 Answers: 3 Comments: 0

lim_(x→1) ((x−1)/(ln((x/(2−x))))) = ???

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{{x}−\mathrm{1}}{{ln}\left(\frac{{x}}{\mathrm{2}−{x}}\right)}\:=\:??? \\ $$

Question Number 142904 Answers: 1 Comments: 0

If 2+log _2 (x)=3+log _3 (y)=log _6 (x−4y) then the value of (1/(2y))−(2/x)=?

$${If}\:\mathrm{2}+\mathrm{log}\:_{\mathrm{2}} \left({x}\right)=\mathrm{3}+\mathrm{log}\:_{\mathrm{3}} \left({y}\right)=\mathrm{log}\:_{\mathrm{6}} \left({x}−\mathrm{4}{y}\right) \\ $$$${then}\:{the}\:{value}\:{of}\:\frac{\mathrm{1}}{\mathrm{2}{y}}−\frac{\mathrm{2}}{{x}}=? \\ $$

Question Number 142765 Answers: 1 Comments: 1

A class has 13 children. To play a game one child is the referee and the other children are divided in three teams with four children in each team. In how many ways can the class play the game?

$${A}\:{class}\:{has}\:\mathrm{13}\:{children}.\:{To}\:{play}\:{a} \\ $$$${game}\:{one}\:{child}\:{is}\:{the}\:{referee}\:{and}\:{the} \\ $$$${other}\:{children}\:{are}\:{divided}\:{in}\:{three}\: \\ $$$${teams}\:{with}\:{four}\:{children}\:{in}\:{each}\:{team}. \\ $$$${In}\:{how}\:{many}\:{ways}\:{can}\:{the}\:{class}\:{play} \\ $$$${the}\:{game}? \\ $$

Question Number 142691 Answers: 1 Comments: 1

The number of distributions of 52 cards divided equally to 4 persons so as each gets 4 cards of same suit taken away from 3suits(4×3=12)℘ remaining card from remaining 4 th suit is

$${The}\:{number}\:{of}\:{distributions}\:{of}\:\mathrm{52} \\ $$$${cards}\:{divided}\:{equally}\:{to}\:\mathrm{4}\:{persons}\:{so} \\ $$$${as}\:{each}\:{gets}\:\mathrm{4}\:{cards}\:{of}\:{same}\:{suit} \\ $$$${taken}\:{away}\:{from}\:\mathrm{3}{suits}\left(\mathrm{4}×\mathrm{3}=\mathrm{12}\right)\wp \\ $$$${remaining}\:{card}\:{from}\:{remaining} \\ $$$$\mathrm{4}\:{th}\:{suit}\:{is} \\ $$

Question Number 142630 Answers: 0 Comments: 1

Given f(x)=(1/(1+2^x )) find the value of f((1/(2018)))×f((3/(2018)))×f(((2015)/(2018)))×f(((2017)/(2018)))=?

$$\:{Given}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}^{{x}} } \\ $$$${find}\:{the}\:{value}\:{of} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2018}}\right)×{f}\left(\frac{\mathrm{3}}{\mathrm{2018}}\right)×{f}\left(\frac{\mathrm{2015}}{\mathrm{2018}}\right)×{f}\left(\frac{\mathrm{2017}}{\mathrm{2018}}\right)=? \\ $$

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