0
Still curious? Ask our experts.
Chat with our AI personalities
Ezra
Taiga
ViviAdd your answer:
Earn +20 pts
Maria keeps her four stuffed bears lined up on a shelf over her bed how many arrangements of the bears are possible?
The answer is 4! (4 factorial), the same as 4x3x2x1, which equals 24 combinations. The answer is 24 and this is how: A b c d A b d c A c d b A c b d A d c b A d b c B c d a B c a d B d a c B d c a B a c d B a d c C d a b C d b a C a b d C a d b C b d a C b a d D a b c D a c b D b c a D b a c D c a b D c b a
C program to print all combinations of a 4-digit number?
#include<stdio.h> int main() { int a,b,c,d; for(a=1; a<5; a++) { for(b=1; b<5; b++) { for(c=1; c<5; c++) { for(d=1; d<5; d++) { if(!(a==b a==c a==d b==c b==d c==d)) printf("dd\n",a,b,c,d); } } } } return 0; }
How can you write a c program to find maximum element in row and minimum element in column of a matrix?
#include <stdio.h> main() { int m, n, c, d, A[10][10],temp=0; printf("\nEnter the number of rows and columns for matrix A:\n"); scanf("%d%d", &m, &n); printf("\nEnter the elements of matrix A:\n"); for ( c = 0 ; c < m ; c++ ) for ( d = 0 ; d < n ; d++ ) scanf("%d", &A[c][d]); printf("\nMatrix entered is:\n"); for ( c = 0 ; c < m ; c++ ) { for ( d = 0 ; d < n ; d++ ) printf("%d\t", A[c][d]); printf("\n"); } printf("\n\nMaximum element of each row is:\n"); for(c=0;c<m;c++) { for(d=0;d<n;d++) { if(A[c][d]>temp) temp=A[c][d]; } printf("\n\t\tRow %d: %d",c+1,temp); temp=0; } temp=A[0][0]; printf("\n\nMinimum element of each coloumn is:\n"); for(c=0;c<n;c++) { for(d=0;d<m;d++) { if(A[d][c]<temp) temp=A[d][c]; } printf("\n\t\tColoumn %d: %d",c+1,temp); temp=A[d][c] ; } return 0; }
A c plus plus program illustrating rational number?
int a=2, b=3, c=4, d=5; printf ("%d/%d + %d/%d = %d/%d\n", a, b, c, d, a*d+b*c, b*d);
If a over b equals c over d then is it possible that a plus b equals c plus d and why?
Yes. To show the conditions on a, b, c and d given that if a/b = c/d then a+b = c+d. Suppose b != d (and that both b and d are non-zero) then: d = kb for some number k (!= 0), so c/d = c/kb = (c/k)/b so a/b = (c/k)/b => a = c/k => c = ka Thus: c + d = ka + kb = k(a + b) Which means that c + d = a + b only if k = 1. Thus if a/b = c/d then a + b = c + d only if a = c and b = d. The condition on b and d both being non-zero prevents the possibility of division by zero. If either is zero, a division by zero will occur and at least one of the fractions is infinite.
