There are three teams `x`,`x+1` and `y` childrens and total number of childrens in the teams is `24`. if two childrens of the same team do not fight, then

Options

A. (a) maximum number of fights is 190
B. (b) maximum number of fights is 191
C. (c) maximum number of fights occur when x=7
D. (d) maximum number of fights occur when x=8

Answer

B, C, D

Explanation

`becausex+x+1+y=24`
`impliesy=23-2x`
Let N=Total number of fights subject to the condition that any two children of one tem do not fight.
`thereforeN=2.^(24)C_(2)-(.^(x)C_(2)+.^(x+1)C_(2)+.^(y)C_(2))`
`=.^(24)C_(2)-(.^(x)C_(2)+.^(x+1)C_(2)+.^(23-2x)C_(2))`
`=23-3x^(2)+45x`
`therefore(dN)/(dx)=0-6x+45`
for maximum or minimum, put `(dN)/(dx)=0impliex=7.5`
`impliesx=7" "[because x in I]`
Now, `(d^(2)N)/(dx^(2)) lt0`
`therefore` N will be maximum when x=7
and N=23-`3(7)^(2)+45xx7=191`