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Permutation and CombinationQuestion and Answers: Page 19

Question Number 56212 Answers: 1 Comments: 0

In how many ways can 4 boys and 3 girls stand in a straight line a. if there are no restrictions b. if the boys stand next to each other

$In how many ways can 4 boys and 3 girls$ $stand in a straight line$ $a . if there are no restrictions$ $b . if the boys stand next to each other$

Question Number 56025 Answers: 2 Comments: 0

Find the probability that a student arranging the letters of the word MATHEMATICS will make all the vowels be together in any arrangement he or she does.

$Find the probability that a student$ $arranging the letters of the word$ $MATHEMATICS will make all the$ $vowels be together in any arrangement$ $he or she does .$

Question Number 55502 Answers: 2 Comments: 1

Question Number 55468 Answers: 1 Comments: 0

Question Number 55192 Answers: 1 Comments: 1

what will be the numbers of zeroes in the expension of a) 100!×25! b) 100!+25! please help

$w h a t w i l l b e t h e n u m b e r s o f z e r o e s i n t h e e x p e n s i o n o f$ $a) 100! \times 25!$ $b) 100! + 25!$ $p l e a s e h e l p$

Question Number 55119 Answers: 1 Comments: 1

How many numbers, divisible by 5, can be made with the digits 2,3,4 and 5 where no digit is being used more than once in each number?

$How many numbers, divisible by 5, can$ $be made with the digits 2, 3, 4 and 5 where$ $no digit is being used more than once$ $in each number ?$

Question Number 55115 Answers: 0 Comments: 4

Question Number 55100 Answers: 1 Comments: 5

Question Number 55099 Answers: 1 Comments: 0

Show that for n ∈ N, Σ_(r=0) ^n P_r ^n = ⌊n! e⌋ where ⌊x⌋ denotes the greatest integer ≤ x and P_r ^n = ((n!)/((n − r)!))

$Show that for n \in N,$ $\underset{r = 0}{\sum^{n}} P_{r}^{n} = ⌊ n! e ⌋$ $where ⌊ x ⌋ denotes the greatest integer ⩽ x$ $and P_{r}^{n} = \frac{n!}{(n - r)!}$

Question Number 54860 Answers: 1 Comments: 1

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

$P r o v e t h a t$ $1 . (\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n - 1 \\ r \end{matrix}) + (\begin{matrix} n - 1 \\ r - 1 \end{matrix})$ $2 . (\begin{matrix} n \\ r \end{matrix}) + (\begin{matrix} n \\ r - 1 \end{matrix}) = (\begin{matrix} n + 1 \\ r \end{matrix})$ $3 . (\begin{matrix} n \\ 0 \end{matrix}) + (\begin{matrix} n \\ 1 \end{matrix}) + (\begin{matrix} n \\ 2 \end{matrix}) + . . + (\begin{matrix} n \\ n \end{matrix}) = 2^{n}$

Question Number 54698 Answers: 2 Comments: 0

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

$H o w m a n y w o r d s w i t h a t l e a s t 2 l e t t e r s$ $c a n b e f o r m e d u s i n g t h e l e t t e r s f r o m$ $T I N K U T A R A ?$

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)

$Such That$ $a ._{n + 1} C_{r} = \frac{(n + 1) ._{n} C_{r}}{(n - r + 1)}$ $b ._{n} C_{0} +_{n} C_{2} +_{n} C_{4 . . .} =_{n} C_{1} +_{n} C_{3} +_{n} C_{5 . . .} = 2^{n - 1}$

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

$I f a l l t h e w o r d s w i t h o r w i t h o u t$ $m e a n i n g a r e w r i t t e n u s i n g t h e$ $l e t t e r s o f t h e w o r d Q U E E N a n d$ $a r e a r r a n g e d a s i n E n g l i s h$ $d i c t i o n a r y . W h a t i s t h e p o s i t i o n$ $o f t h e w o r d Q U E E N ?$ $a) 4 4 t h b) 45 t h c) 46 t h d) 47 t h$

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

$T h e s u m o f t h e d i g i t s i n t h e u n i t$ $p l a c e o f a l l f o u r d i g i t n u m b e r s$ $f o r m e d u s i n g t h e n u m b e r s 3 4 5 a n d L$ $6 w i t h o u t r e p e t i t i o n i s$ $a) 4 32 b) 108 c) 36 d) 18$

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

$8 d i g i t n u m b e r s a r e f o r m e d u s i n g$ $1 1 2 2 2 3 4 4 . T h e n u m b e r o f s u c h$ $n u m b e r s i n w h i c h t h e o d d d i g i t s$ $d o n o t o c c u p y o d d p l a c e s i s$ $a) 1 60 b) 120 c) 60 d) 48$

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

$A n 8 d i g i t n u m b e r d i v i s i b l e b y 9 i s$ $t o b e f o r m e d u s i n g d i g i t s f r o m$ $0 t o 9 w i t h o u t r e p e a t i n g t h e d i g i t s .$ $T h e n u m b e r o f w a y s i n w h i c h t h i s$ $c a n b e d o n e i s ?$ $a) 7 2 (7 i) b) 18 (7 i) c) 40 (7 i) d) 36 (7 i)$ $p l e a s e I m e a n f a c t o r i a l b y i . T h e$ $p a r t o f m y s c r e e n w h e r e t h e f a c t o r i a l$ $s y m b o l i s i s n t f u n c t i o n i n g . T h a n k s$

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...

$Q = I n h o w m a n y w a y s 8 s t u d e n t s c a n b e d i v i d e d i n t o t w o e q u a l g r o u p s ?$ $A n s = \frac{8!}{2! \times {(4!)}^{2}}$ $h e r e w h y t w o f a c t o r i a l i s d i v i d e d ?$ $w e d o n o t d o t h e s a m e i n c a s e o f d i s t r i b u t i n g 52 c a r d s e q u a l l y a m o n g 4 p e r s o n s .$ $p l s e x p l a i n . . .$

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

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

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?

$A d e l e g a t i o n o f 4 s t u d e n t s i s t o b e$ $s e l e c t e d f r o m a t o t a l o f 12 s t u d e n t s .$ $I n h o w m a n y w a y s c a n t h e d e l e g a t i o n$ $b e s e l e c t e d i f :$ $1) 2 p a r t i c u l a r s t u d e n t s w i s h t o b e$ $i n c l u d e d t o g e t h e r o n l y i n t h e d e l e g a t i o n ?$ $2) 2 p a r t i c u l a r s t u d e n t s r e f u s e t o b e$ $t o g e t h e r a n d 2 o t h e r p a r t i c u l a r s t u d e n t s$ $w i s h t o b e t o g e t h e r o n l y i n t h e d e l e g a t i o n ?$

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 ?

$5 - d i g i t n u m b e r d i v i s i b l e b y 3 f o r m e d$ $u s i n g 0, 1, 2, 3, 4, 5 w i t h o u t r e p e t i t i o n .$ $T o t a l n u m b e r o f s u c h n o . s i s ?$

Pg 14 Pg 15 Pg 16 Pg 17 Pg 18 Pg 19 Pg 20 Pg 21 Pg 22 Pg 23

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