106okk thanx sir..
The number of non- negative integral solutions of x1+x2+x3+x4=20 with at least one of the xi ( i=1,2,3,4) is even is:
(a) 1500
(b) 1470
(c) 1450
(d) 1606
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29 Answers
1Yaar....agar 12th class ke ladkon ko sir bulaagoge...to budhe aadmi-o ko kya bologer ???? x-(
49@megan: consider this case:
suppose an equation is gven thus: x + y = 9 and u have to find number of integral solutions..
it means u have to find no of the possible solutions in which x and y are integers i.e. 0,1,2,3,-1,-7,....etc...
and we have not to include cases where x and y are fractions...like case where x=4.5 and y=4.5 will not be considered!! :)
hope this clears ur doubt!!
@bicchuram: LOL!! u r being called sir these days! :P
1@bicchuram : sir cud u plz tell me as to what are integral solutions??? this might sound silly but appreciate ur help
1Sir not my doubt,....thought this wud interest many on this site
:-)
1Then this is a quest. from arihant :
If a,b,c are odd +ve integers....then no. of integral solns. of a+b+c= 13 is ____________.
Ans. : 21
1So an indeterminate eqn having infinite no. of solns......??
that's true.
62take any x1, x2 and x3
then x4=20-x1-x2-x3
so for any values of x1, x2 and x3 .. you have a x4
1Seen some question like it b4...so just trying to solve it....neways...dunno the answer.....just busy doing it....
Sir how did u find it infinity ???
1Yup ....
x1+x2+x3+x4 = 20
Solns. = 23C3
all odds :
x1 = 2y1 + 1 (all odds of form 2m+1)
Hence 2(y1+y2+y3+y4) + 4 = 20
y1+y2+y3+y4 = 8
solns. with odd +ve integers .... = 11C3
Hence at least one even : 23C3 - 11C3 = 1606
ANOTHER ADDITION TO THIS QUEST WHICH MAY MAKE IT INTERESTING :
FIND ALL INTEGRAL SOLNS> WITH AT LEAST ONE EVEN NO> (REMEMBER ALL SOLNS INVOLVE BOTH +ve and -ve SOLN.
62total solutions is 20+3C3 = 23x22x21/6 = 23*11*7 = 161*11 = 1771
No of solutions with all odds
2(y1+y2+y3+y4) + 4 = 20
(y1+y2+y3+y4) = 8
so it is 11C3 = 11*10*9/6 = 165
The difference is 1606
[1]
62@asish:
Quote:
So, 2k1+2k2+(2k3+1)+(2k4+1)=20
=> k1+k2+k3+k4 = 9
=> no. of ways = 12C3
here you have to multiply this by 4C2
because 1st 2 could be even, 1st and 3rd could be even, 1st/4th, 2nd/3rd, 2nd/4th, 3rd/4th...
62@ eureka..
THat step is basically because if all xi's are odd then
xi=2yi+1 for each xi
so
x1+x2+x3+x4=(2y1+1)+(2y2+1)+(2y3+1)+(2y4+1) = 2(y1+y2+y3+y4) + 4 = 20
62@soumi.. the method is this
there are 20 numbers which you have to divide in 4 parts
so you are effectively putting 3 dividers
Hence you have to arrrange 20 numbers using 3 dividers
if the numbers are all 1, they are identical.. the dividers are also identical
so the question reduces to arranging 3 dividers of one kind and 20 numbers of one kind in a row
which is 23! / (3! x 20!)
106yeah realised that .... and edited...
actually dont have pen and paper with me rite now..
62=> k1+k2+k3+k4=10
=> no. of ways = 14C3
asish this is wrong: no of ways is 13C3 not 14 C 3
The other one also needs to be fixed... to 12 C 3
1Y is the total no. of solutions 20+3 c3?? (it may b a silly question!!!!, but i really dont know!!)
106eure, let xi = 2ki + 1 where ki ≥0
Put in the equation Σxi = 20
106if atleast one of xi is even,
then the possible cases are:
all even, 2even 2odd ....
If all even then
2k1 + 2k2+2k3+2k4=20
=> k1+k2+k3+k4=10
=> no. of ways = 13C3
If two even and two odd
So, 2k1+2k2+(2k3+1)+(2k4+1)=20
=> k1+k2+k3+k4 = 9
=> no. of ways = 12C3
so total = 13C3+12C3