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Permutation and CombinationQuestion and Answers: Page 6

Question Number 178937 Answers: 1 Comments: 0

Find the greatest coefficient in expansion of: (6 − 4x)^(− 3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{expansion}\:\mathrm{of}:\:\:\:\:\left(\mathrm{6}\:\:\:−\:\:\:\mathrm{4x}\right)^{−\:\:\mathrm{3}} \\ $$

Question Number 178454 Answers: 2 Comments: 0

In a chess board number of unit squares with 1)one vertex common? 2)2 vertices common?? 3)2 sides common??

$${In}\:{a}\:{chess}\:{board}\:{number}\:{of}\:{unit}\:{squares} \\ $$$$\left.{with}\:\mathrm{1}\right){one}\:{vertex}\:{common}? \\ $$$$\left.\mathrm{2}\right)\mathrm{2}\:{vertices}\:{common}?? \\ $$$$\left.\mathrm{3}\right)\mathrm{2}\:{sides}\:{common}?? \\ $$

Question Number 178395 Answers: 2 Comments: 0

How many 5 digit numbers with different digits are multiple of 9?

$${How}\:{many}\:\mathrm{5}\:{digit}\:{numbers}\:{with} \\ $$$${different}\:{digits}\:{are}\:{multiple}\:{of}\:\mathrm{9}? \\ $$

Question Number 178173 Answers: 2 Comments: 0

Question Number 177597 Answers: 2 Comments: 0

Question Number 177309 Answers: 1 Comments: 1

How many 3 digited numbers which are divisible by 1)3 2)4 3)5 4)6 5)7 6)8 7)9 with repetetion of digits is NOT allowed...one problem process

$${How}\:{many}\:\mathrm{3}\:{digited}\:{numbers}\: \\ $$$$\left.{which}\:{are}\:{divisible}\:{by}\:\mathrm{1}\right)\mathrm{3} \\ $$$$\left.\mathrm{2}\left.\right)\left.\mathrm{4}\left.\:\left.\:\left.\:\:\mathrm{3}\right)\mathrm{5}\:\:\:\:\mathrm{4}\right)\mathrm{6}\:\:\:\mathrm{5}\right)\mathrm{7}\:\:\:\:\mathrm{6}\right)\mathrm{8}\:\:\mathrm{7}\right)\mathrm{9} \\ $$$${with}\:{repetetion}\:{of}\:{digits}\:{is} \\ $$$${NOT}\:{allowed}...{one}\:{problem}\:{process} \\ $$

Question Number 177040 Answers: 1 Comments: 0

Question Number 177008 Answers: 1 Comments: 0

Question Number 177007 Answers: 1 Comments: 0

Question Number 175567 Answers: 0 Comments: 0

in how many ways can you put 40 identical balls into 20 identical boxes such that each box obtains at least one ball and at most 5 balls?

$${in}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{put}\:\mathrm{40} \\ $$$${identical}\:{balls}\:{into}\:\mathrm{20}\:{identical}\:{boxes} \\ $$$${such}\:{that}\:{each}\:{box}\:{obtains}\:{at}\:{least}\:{one} \\ $$$${ball}\:{and}\:{at}\:{most}\:\mathrm{5}\:{balls}? \\ $$

Question Number 174855 Answers: 1 Comments: 0

Find the number of ways a committee of 4 people can be chosen from a group of 5 men and 7 women when it contains people of both sexes and there are at least as many women as men.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\: \\ $$$$\mathrm{a}\:\mathrm{committee}\:\mathrm{of}\:\mathrm{4}\:\mathrm{people}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{from}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{5}\:\mathrm{men} \\ $$$$\mathrm{and}\:\mathrm{7}\:\mathrm{women}\:\mathrm{when}\:\mathrm{it}\:\mathrm{contains} \\ $$$$\mathrm{people}\:\mathrm{of}\:\mathrm{both}\:\mathrm{sexes}\:\mathrm{and} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least}\:\mathrm{as}\:\mathrm{many} \\ $$$$\mathrm{women}\:\mathrm{as}\:\mathrm{men}. \\ $$

Question Number 173976 Answers: 1 Comments: 0

B(a,b)=∫_0 ^1 x^(a−1) (1−x)^(b−1) dx Γ(s)= ∫_0 ^∞ t^(s−1) e^(−t) dt Why B(a,b)= ((Γ(a)Γ(b))/(Γ(a+b))) ?

$$ \\ $$$$\:\:\:\:{B}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx}\: \\ $$$$\:\:\:\:\:\:\:\Gamma\left({s}\right)=\:\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$ \\ $$$$\:\:{Why}\:\:\:\:{B}\left({a},{b}\right)=\:\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}\:? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 173975 Answers: 0 Comments: 0

∫_0 ^∞ (y^(a−1) /((1+y)^b )) dy =^(u=(1/(1+y))) ∫_0 ^1 (((1−u)^(a−1) )/u^(a−1) ) u^b (du/u^2 ) = ∫_0 ^1 u^(b−a−1) (1−u)^(a−1) du = B(b−a,a)=((Γ(b−a)Γ(a))/(Γ(b)))

$$ \\ $$$$\: \\ $$$$ \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{y}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{y}\right)^{{b}} }\:{dy}\:\overset{{u}=\frac{\mathrm{1}}{\mathrm{1}+{y}}} {=}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{u}\right)^{{a}−\mathrm{1}} }{{u}^{{a}−\mathrm{1}} }\:{u}^{{b}} \:\:\frac{{du}}{{u}^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{{b}−{a}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{a}−\mathrm{1}} {du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{B}\left({b}−{a},{a}\right)=\frac{\Gamma\left({b}−{a}\right)\Gamma\left({a}\right)}{\Gamma\left({b}\right)} \\ $$$$\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 173354 Answers: 0 Comments: 2

A three−digit odd number less than 500 is to be formed from 1,2,3,4 and 5. If repetition of digits is allowed, in how many ways can this be done?

$$\mathrm{A}\:\mathrm{three}−\mathrm{digit}\:\mathrm{odd}\:\mathrm{number}\:\mathrm{less}\:\mathrm{than}\:\mathrm{500} \\ $$$$\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\mathrm{and}\:\mathrm{5}.\:\mathrm{If}\:\mathrm{repetition} \\ $$$$\mathrm{of}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{allowed},\:\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways} \\ $$$$\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{done}? \\ $$

Question Number 173014 Answers: 0 Comments: 0

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Question Number 172817 Answers: 2 Comments: 0

Question Number 172356 Answers: 0 Comments: 0

Question Number 172354 Answers: 1 Comments: 0

Question Number 172347 Answers: 0 Comments: 0

Question Number 172008 Answers: 0 Comments: 0

solve: 15(2n)_C_((n−1)) =28(2n−1)_C_n . find n

$${solve}: \\ $$$$\mathrm{15}\left(\mathrm{2}{n}\right)_{{C}_{\left({n}−\mathrm{1}\right)} } =\mathrm{28}\left(\mathrm{2}{n}−\mathrm{1}\right)_{{C}_{{n}} } .\:{find}\:{n} \\ $$

Question Number 171845 Answers: 0 Comments: 0

n_c_(r+1) + n_c_r =n+1_c_(r .solve for n and r.)

$${n}_{{c}_{{r}+\mathrm{1}} } +\:{n}_{{c}_{{r}} } ={n}+\mathrm{1}_{{c}_{{r}\:\:.{solve}\:{for}\:\:{n}\:{and}\:{r}.} } \\ $$

Question Number 171761 Answers: 0 Comments: 0

solve: ((sin(10+x))/(sin(13+x)))=((sin27sin39)/(sin24sin57))

$${solve}: \\ $$$$\frac{{sin}\left(\mathrm{10}+{x}\right)}{{sin}\left(\mathrm{13}+{x}\right)}=\frac{{sin}\mathrm{27}{sin}\mathrm{39}}{{sin}\mathrm{24}{sin}\mathrm{57}} \\ $$

Question Number 171223 Answers: 0 Comments: 0

In how many ways can you select 4 from 40 persons if every two persons may be selected together at most one time?

$${In}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{select}\:\mathrm{4} \\ $$$${from}\:\mathrm{40}\:{persons}\:{if}\:{every}\:{two}\:{persons} \\ $$$${may}\:{be}\:{selected}\:{together}\:{at}\:{most}\:{one}\: \\ $$$${time}? \\ $$

Question Number 171012 Answers: 1 Comments: 1

The number of five digits can be made with the digits 1, 2, 3 each of which can be used atmost thrice in a number is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{five}\:\mathrm{digits}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}\:\mathrm{each}\:\mathrm{of}\:\mathrm{which}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{used}\:\mathrm{atmost}\:\mathrm{thrice}\:\mathrm{in}\:\mathrm{a}\:\mathrm{number}\:\mathrm{is} \\ $$

Question Number 171011 Answers: 2 Comments: 0

How many 5-digit numbers from the digits {0, 1, ....., 9} have? (i) Strictly increasing digits (ii) Strictly increasing or decreasing digits (iii) Increasing digits (iv) Increasing or decreasing digits

$$\mathrm{How}\:\mathrm{many}\:\mathrm{5}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{digits}\:\left\{\mathrm{0},\:\mathrm{1},\:.....,\:\mathrm{9}\right\}\:\mathrm{have}? \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Strictly}\:\mathrm{increasing}\:\mathrm{digits} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Strictly}\:\mathrm{increasing}\:\mathrm{or}\:\mathrm{decreasing} \\ $$$$\mathrm{digits} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{Increasing}\:\mathrm{digits} \\ $$$$\left(\mathrm{iv}\right)\:\mathrm{Increasing}\:\mathrm{or}\:\mathrm{decreasing}\:\mathrm{digits} \\ $$

Question Number 170794 Answers: 2 Comments: 0

2n objects of each of three kinds are given to two persons, so that each person gets 3n objects. Prove that this can be done in 3n^2 + 3n + 1 ways.

$$\mathrm{2}{n}\:\mathrm{objects}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{three}\:\mathrm{kinds}\:\mathrm{are} \\ $$$$\mathrm{given}\:\mathrm{to}\:\mathrm{two}\:\mathrm{persons},\:\mathrm{so}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{person}\:\mathrm{gets}\:\mathrm{3}{n}\:\mathrm{objects}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{1}\:\mathrm{ways}. \\ $$

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