1SOMEBODY PLEASE HELP !!!!!!!!!!!!!!!!!!!!!
Find the no. of integer solutions of x+y+z = 24 such that x is even and +ve , y>2 and z > -2 ?
A) 110
B) 96
C) 120
D) 121
-
UP 0 DOWN 0 1 9
9 Answers
4@∫ : Great soln but how cum the ans not in options
Aveek can u verify the answer once
1PROPHET SIR ... PLEASE TELL ME WHERE I'VE GONE WRONG IN MY SOLN :
IF X IS EVEN Y AND Z BOTH HAVE TO BE EITHER ODD OR EVEN.
IF BOTH ARE EVEN THEN :
2M + 2N + 2L = 24 WHERE M>0 N>1 AND L>-1
OR 2M + 2(N-1) + 2(L+1) = 24
OR M + N + L = 12 SO SOLN = 11C2
IF BOTH ARE ODD
2M + (2N+1) + (2L+1) = 24 2N+1>2 2L+1>-2
SO WE CAN ONCE AGAIN FIND THE NO. OF SOLNS. FROM HERE.
AND IF WE ADD UP THE2 CASES....WE SHOULD GET THE ANSWER.....WHICH I CAN'T DO.....
PROPHET SIR PLEASE HELP
1ELEMENTARY SOLN:l
@ Aveek
i tried this problem initially using Multinomial theorem..but was stuck there..so thought of some simpler solution..
we have three variables here x shud be even ... y > 2 and z > -2
so lets fix a value for x as 2..we will get 21 values for y and z..
now fix x = 4 we will get 19 values for x and y
similarly this series will continue
so we get number of solutions = 21 +19 +17.....+1 = 121
:)