Answer Posted / vadivelt

My Assumption would be.,

N = Some no
P = Some no
NP -> does not mean multiplication of N and P. Just it
means two different numbers.

So my answer is.,
4! = 4*3*2*1

ie.,
4! = 24 so,
P! = NP
Hence proved..

Here P = 4, N = 2.

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1263

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I need testPalindrome and removeSpace #include #define SIZE 256 /* function prototype */ /* test if the chars in the range of [left, right] of array is a palindrome */ int testPalindrome( char array[], int left, int right ); /* remove the space in the src array and copy it over to the "copy" array */ /* set the number of chars in the "copy" array to the location that cnt points t */ void removeSpace(char src[], char copy[], int *cnt); int main( void ) { char c; /* temporarily holds keyboard input */ char string[ SIZE ]; /* original string */ char copy[ SIZE ]; /* copy of string without spaces */ int count = 0; /* length of string */ int copyCount; /* length of copy */ printf( "Enter a sentence:\n" ); /* get sentence to test from user */ while ( ( c = getchar() ) != '\n' && count < SIZE ) { string[ count++ ] = c; } /* end while */ string[ count ] = '\0'; /* terminate string */ /* make a copy of string without spaces */ removeSpace(string, copy, ©Count); /* print whether or not the sentence is a palindrome */ if ( testPalindrome( copy, 0, copyCount - 1 ) ) { printf( "\"%s\" is a palindrome\n", string ); } /* end if */ else { printf( "\"%s\" is not a palindrome\n", string ); } /* end else */ return 0; /* indicate successful termination */ } /* end main */ void removeSpace(char src[], char copy[], int *cnt) { } int testPalindrome( char array[], int left, int right ) { }

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