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Permutation and CombinationQuestion and Answers: Page 14

Question Number 108450 Answers: 2 Comments: 2

((BeMath)/(⊂⊃)) (1)find ((1/2))! (2)∫_0 ^(π/2) ((x sin x)/((1+cos x)^2 )) dx

$\frac{B e M a t h}{\subset\supset}$ $(1) f i n d (\frac{1}{2})!$ $(2) \underset{0}{\int^{π / 2}} \frac{x \sin x}{{(1 + \cos x)}^{2}} d x$

Question Number 108262 Answers: 2 Comments: 0

∫_0 ^1 (y^y )^((y^y )^((y^y )) ) dy=? please help

$\int_{0}^{1} {(y^{y})}^{{(y^{y})}^{(y^{y})}} dy = ?$ $please help$

Question Number 107844 Answers: 0 Comments: 0

A general case: we have totally n letters, among them n_1 times A, n_2 times B, n_3 times C, n_4 times D etc. (n_1 ,n_2 ,n_3 ,n_(4,) ...≥2, n>n_1 +n_2 +n_3 +n_4 +....) how many different words can be formed using these n letters such that same letters are not next to each other. see also Q107451.

$A g e n e r a l c a s e :$ $w e h a v e t o t a l l y n l e t t e r s, a m o n g t h e m$ $n_{1} t i m e s A, n_{2} t i m e s B, n_{3} t i m e s C,$ $n_{4} t i m e s D e t c .$ $(n_{1}, n_{2}, n_{3}, n_{4,} . . . ⩾ 2, n > n_{1} + n_{2} + n_{3} + n_{4} + . . . .)$ $h o w m a n y d i f f e r e n t w o r d s c a n b e$ $f o r m e d u s i n g t h e s e n l e t t e r s s u c h t h a t$ $s a m e l e t t e r s a r e n o t n e x t t o e a c h$ $o t h e r .$ $s e e a l s o Q 107451 .$

Question Number 107451 Answers: 1 Comments: 1

How many words can you form using the letters in UNUSUALLY such that no same letters are next to each other? [Answer: 10200]

$H o w m a n y w o r d s c a n y o u f o r m u s i n g$ $t h e l e t t e r s i n U N U S U A L L Y$ $s u c h t h a t n o s a m e l e t t e r s a r e n e x t$ $t o e a c h o t h e r ?$ $[A n s w e r : 10200]$

Question Number 107327 Answers: 0 Comments: 0

1≤p≤k≤n show that Σ_(k=p) ^n C_(k−1) ^(p−1) =C_n ^p please i need help

$1 ⩽ p ⩽ k ⩽ n$ $s h o w t h a t \underset{k = p}{\sum^{n}} C_{k - 1}^{p - 1} = C_{n}^{p}$ $p l e a s e i n e e d h e l p$

Question Number 106815 Answers: 1 Comments: 0

please help me to show that the equation X^n +aX+c=0 can not have more than 3 reals solutions

$p l e a s e h e l p m e t o s h o w$ $t h a t t h e e q u a t i o n$ $X^{n} + a X + c = 0 c a n n o t h a v e$ $m o r e t h a n 3 r e a l s s o l u t i o n s$

Question Number 106792 Answers: 1 Comments: 0

≻bobhans≺ From a batch containing 6 boys and 4 girls a group of 4 students is tobe selected . How many group formations will have exactly 2 girls?

$≻ bobhans ≺$ $From a batch containing 6 boys and 4 girls$ $a group of 4 students is tobe selected .$ $How many group formations will have$ $exactly 2 girls ?$

Question Number 106305 Answers: 1 Comments: 0

what is probability of 5 coming up at least one if a die is rolled 3 times

$what is probability of 5 coming up at least one if a die$ $is rolled 3 times$

Question Number 106292 Answers: 1 Comments: 0

a box contains 4 blue, 3 green and 2 red identicall balls. if two balls are selected at random without replacement , what is the probability that two balls be of the same colours?

$a box contains 4 blue, 3 green and 2 red identicall balls .$ $if two balls are selected at random without$ $replacement, what is the probability$ $that two balls be of the same colours ?$

Question Number 105994 Answers: 1 Comments: 1

Given { ((2log _5 (x+2)−log _5 (y)+(1/2)=0)),((log _5 (x^2 +2x−2)=0 )) :} find the value of y

$G i v e n {\begin{cases} 2 \log_{5} (x + 2) - \log_{5} (y) + \frac{1}{2} = 0 \\ \log_{5} (x^{2} + 2 x - 2) = 0 \end{cases}$ $f i n d t h e v a l u e o f y$

Question Number 105853 Answers: 0 Comments: 5

A box contains 5 balls, 2 balls were drawn at random, both of which turned out tobe white. what is the probability that all the balls in the box are white ?

$A b o x c o n t a i n s 5 b a l l s, 2 b a l l s w e r e$ $d r a w n a t r a n d o m, b o t h o f w h i c h$ $t u r n e d o u t t o b e w h i t e . w h a t i s t h e$ $p r o b a b i l i t y t h a t a l l t h e b a l l s i n t h e b o x$ $a r e w h i t e ?$

Question Number 105791 Answers: 1 Comments: 1

From 1 to 12345, how many numbers contain the digit 0? Find the number of zeros in all these numbers. Example: 10020 has three zeros.

$F r o m 1 t o 12345, h o w m a n y n u m b e r s$ $c o n t a i n t h e d i g i t 0 ? F i n d t h e n u m b e r$ $o f z e r o s i n a l l t h e s e n u m b e r s .$ $E x a m p l e : 10020 h a s t h r e e z e r o s .$

Question Number 105551 Answers: 2 Comments: 0

In how many different ways can 10 students be divided into 3 groups?

$I n h o w m a n y d i f f e r e n t w a y s c a n 10$ $s t u d e n t s b e d i v i d e d i n t o 3 g r o u p s ?$

Question Number 105265 Answers: 3 Comments: 0

Question Number 104370 Answers: 1 Comments: 0

Question Number 104155 Answers: 0 Comments: 3

Question Number 104134 Answers: 2 Comments: 0

what is the coefficient x^(15) in the expansion of x^6 (1−x)^(11)

$w h a t i s t h e c o e f f i c i e n t x^{15}$ $i n t h e e x p a n s i o n o f x^{6} {(1 - x)}^{11}$

Question Number 103903 Answers: 1 Comments: 5

a 100 cm long rod should be divided into 3 parts. the length of each part in cm should be integer. in how many different ways can this be done?

$a 100 c m l o n g r o d s h o u l d b e d i v i d e d$ $i n t o 3 p a r t s . t h e l e n g t h o f e a c h p a r t$ $i n c m s h o u l d b e i n t e g e r . i n h o w$ $m a n y d i f f e r e n t w a y s c a n t h i s b e$ $d o n e ?$

Question Number 103826 Answers: 1 Comments: 0

In the expansion of (1+x)^(20) if the coefficient of x^r is twice the coefficient of x^(r−1) , what the value of the coefficient?

$I n t h e e x p a n s i o n o f {(1 + x)}^{20} i f t h e$ $c o e f f i c i e n t o f x^{r} i s t w i c e t h e c o e f f i c i e n t$ $o f x^{r - 1}, w h a t t h e v a l u e o f t h e$ $c o e f f i c i e n t ?$

Question Number 103716 Answers: 1 Comments: 3

Question Number 103606 Answers: 3 Comments: 1

an integer n between 1 and 98 , inclusive is to be chosen at random. what is the probability that n(n+1) will be divisible by 3

$a n i n t e g e r n b e t w e e n 1 a n d 98,$ $i n c l u s i v e i s t o b e c h o s e n a t$ $r a n d o m . w h a t i s t h e p r o b a b i l i t y$ $t h a t n (n + 1) w i l l b e d i v i s i b l e b y 3$

Question Number 103513 Answers: 2 Comments: 0

coefficient of x^5 in expansion (1+x^2 )^5 (1+x)^4 equal to (a) 40 (b) 45 (c) 50 (d) 55 (e) 60

$c o e f f i c i e n t o f x^{5} i n e x p a n s i o n$ ${(1 + x^{2})}^{5} {(1 + x)}^{4} e q u a l t o$ $(a) 40 (b) 45 (c) 50 (d) 55$ $(e) 60$

Question Number 103410 Answers: 1 Comments: 0

Σ_(k=0) ^n (((−1)^k )/(k!))=?

$\underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{k!} = ?$

Question Number 103406 Answers: 0 Comments: 3

100 students are standing in a row. 4 students should be selected in the way that no two of them are next to each other. how many ways do you have to do this?

$100 s t u d e n t s a r e s t a n d i n g i n a r o w .$ $4 s t u d e n t s s h o u l d b e s e l e c t e d i n t h e$ $w a y t h a t n o t w o o f t h e m a r e n e x t t o$ $e a c h o t h e r . h o w m a n y w a y s d o y o u$ $h a v e t o d o t h i s ?$

Question Number 103367 Answers: 1 Comments: 0

many words with 4 letters can be formed using letters from the word TINKUTARA is ___

$m a n y w o r d s w i t h 4 l e t t e r s c a n b e f o r m e d$ $u s i n g l e t t e r s f r o m t h e w o r d$ $T I N K U T A R A i s___$

Question Number 103357 Answers: 1 Comments: 0

There are 21 students will be trained with 6 trainers available. Every students is trained by 1 coach and every coach trains students with different amounts. many ways of grouping students who will be trained are .... (a) ((21!)/(6!)) (b) 6!.21! (c) ((21!)/(2!.3!.4!.5! )) (d) 6×21! (e) ((21!)/(15!))×6!

$T h e r e a r e 21 s t u d e n t s w i l l b e t r a i n e d$ $w i t h 6 t r a i n e r s a v a i l a b l e . E v e r y s t u d e n t s$ $i s t r a i n e d b y 1 c o a c h a n d e v e r y c o a c h$ $t r a i n s s t u d e n t s w i t h d i f f e r e n t a m o u n t s .$ $m a n y w a y s o f g r o u p i n g s t u d e n t s w h o$ $w i l l b e t r a i n e d a r e . . . .$ $(a) \frac{21!}{6!} (b) 6! . 21! (c) \frac{21!}{2! . 3! . 4! . 5!}$ $(d) 6 \times 21! (e) \frac{21!}{15!} \times 6!$

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