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Permutation and CombinationQuestion and Answers: Page 9

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

Question Number 149439 Answers: 1 Comments: 1

Question Number 149437 Answers: 1 Comments: 1

Question Number 148835 Answers: 1 Comments: 0

Question Number 148092 Answers: 1 Comments: 0

Question Number 147914 Answers: 0 Comments: 0

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

Question Number 147654 Answers: 0 Comments: 1

Question Number 147524 Answers: 0 Comments: 5

Question Number 146824 Answers: 1 Comments: 0

Question Number 145944 Answers: 1 Comments: 1

Question Number 145863 Answers: 1 Comments: 0

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)))=?

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

Question Number 142618 Answers: 0 Comments: 2

In how many ways can committee of 5 be formed from a group of 11 people consisting of 4 teachers and 7 students if there is no restriction in the selection ? _______________________

$$\:\:\:{In}\:{how}\:{many}\:{ways}\:{can}\:{committee} \\ $$$${of}\:\mathrm{5}\:{be}\:{formed}\:{from}\:{a}\:{group}\: \\ $$$${of}\:\mathrm{11}\:{people}\:{consisting}\:{of}\:\mathrm{4}\:{teachers} \\ $$$${and}\:\mathrm{7}\:{students}\:{if}\:{there}\:{is}\:{no}\: \\ $$$${restriction}\:{in}\:{the}\:{selection}\:? \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$

Question Number 142305 Answers: 1 Comments: 0

Question Number 141777 Answers: 0 Comments: 1

A pack of 52 cards distributed equally to 4 people so as 4 cards each from same suit( of any 3 suit=4×3=12) and last card from 4th remaining suit . Number of such distributions is?

$${A}\:{pack}\:{of}\:\mathrm{52}\:{cards}\:{distributed}\:{equally}\:{to} \\ $$$$\mathrm{4}\:{people}\:{so}\:{as}\:\mathrm{4}\:{cards}\:{each}\:{from}\: \\ $$$${same}\:{suit}\left(\:{of}\:{any}\:\mathrm{3}\:{suit}=\mathrm{4}×\mathrm{3}=\mathrm{12}\right) \\ $$$${and}\:{last}\:{card}\:{from}\:\mathrm{4}{th}\:{remaining} \\ $$$${suit}\:.\:{Number}\:{of}\:{such} \\ $$$${distributions}\:{is}? \\ $$$$ \\ $$$$ \\ $$

Question Number 141435 Answers: 1 Comments: 0

Find all the arraangment of all the letters of the word SYLLABUSES such that each word contains the word BUS.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{arraangment}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{letters} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{word}\:\mathrm{SYLLABUSES}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{each}\:\mathrm{word}\:\mathrm{contains}\:\mathrm{the}\:\mathrm{word}\:\mathrm{BUS}. \\ $$

Question Number 139352 Answers: 0 Comments: 1

Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Pg 10 Pg 11 Pg 12 Pg 13

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