Question

need help .

Answer 1

http://prntscr.com/7064ni

Answer 2

@ganeshie8

Answer 3

@iambatman @e.mccormick @dan815

Answer 4

can u plz explain ?

Answer 5

but its given that n1<n2<n3<n4<n5

Answer 6

@perl here

Answer 7

@dan815

Answer 8

Oh, maybe we should tag an expert who seems to answer many questions in mathematics @Nnesha

Answer 9

i did abbot bt no response

Answer 10

n1=2

n2=3
n3=4
n4=5
n5=6

Answer 11

no rvc

Answer 12

@Luigi0210

Answer 13

@nincompoop

Answer 14

@Michele_Laino

Answer 15

@Michele_Laino

Answer 16

@alekos

Answer 17

http://gamutech.com
We'll help with questions you have there and there's a chat no lag like OS and you won't get suspended

Answer 18

@butterflydreamer

Answer 19

Hey

Answer 20

how can I help You

Answer 21

hi.here is the question http://prntscr.com/7064ni

Answer 22

aw yea homie this question

Answer 23

what grade is this for

Answer 24

12th grade .

Answer 25

okay ill show u a pattern and u can try to figure out some general stuff

Answer 26

ok.@dan815

Answer 27

1+2+3+4+10=20
1+2+3+5+9
1+2+3+6+8

now u are onto the 3rd digit

1+2+4+5+8
1+2+4+6+7
..
.
.

Answer 28

i got it .thanks to all

Answer 29

everytime u raise a digit theres more restrictions coming in on the digits to the right

Answer 30

how about considering something like this

Answer 31

in order to find more logical methods to solving this

a+b+c = 10


lets say the only restriction we impose is that a b c are not equal to each other, then there is one way of arrange a b c such that they are arranged in increasing to decreasing order

so total number of ways divided by 3!

Answer 32

I think solving the case where a=/=b=/=c is a simpler case

Answer 33

where a b c are allwoed to be equal, then it can be solved easily with stars and bars method, now we need a good way to see what we can do about the case where a b c are not equal to each other

Answer 34

there seems to be something wrong with that method using stars and bars

Answer 35

total no of ways is 7 .

Answer 36

we cant use stars and bars im pretty sure, unless we do it for the very basic case where a b c are allwoed to equal each other, and we can check to see the total possibilties available to us

Answer 37

\[a+b+c+d+e=20\]
has \(\binom{19}{4}\) solutions in "positive integers"
but \(5!\) does not divide \(\binom{19}{4}\)

Answer 38

Ahh we need to account for duplicates is it

Answer 39

ooh im not sure u can do 10 choose 4 like that , thats illegal because that actually allows 0s in there

Answer 40

i think the way to apply stars and bars with 4 separations in this case is umm slightly diff lemme try to see