given with a positive integer value let’s say ‘val’ and the task is to print the value of binomial coefficient b(n, k) where, n and k be any value between 0 to val and hence display the result.
Binomial coefficient (n, k) is the order of choosing ‘k’ results from the given ‘n’ possibilities. The value of binomial coefficient of positive n and k is given by
$$C_k^n=\frac{n!}{(n-k)!k!}$$
where, n >= k
Input-: B(9,2) Output-:
$$B_2^9=\frac{9!}{(9-2)!2!}$$
$$\frac{9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{6\times 5\times 4\times 3\times 2\times 1)\times 2\times 1}=\frac{362,880}{1440}=252$$
The Binomial Coefficient Table is formed for calculating the multiple values that can be generated between n and k.
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0
Input-: value = 5 Output-:

Approach used in the below program is as follows −
Apply the formula given, if n and k is not 0
B(m, x) = B(m, x - 1) * (m - x + 1) / x
START
Step 1-> declare function for binomial coefficient table
int bin_table(int val)
Loop For int i = 0 and i <= val and i++
print i
Declare int num = 1
Loop For int j = 0 and j <= i and j++
If (i != 0 && j != 0)
set num = num * (i - j + 1) / j
End
print num
End
print </p><p>
Step 2-> In main()
Declare int value = 5
call bin_table(value)
STOP#include <stdio.h>
// Function for binomial coefficient table
int bin_table(int val) {
for (int i = 0; i <= val; i++) {
printf("%2d", i);
int num = 1;
for (int j = 0; j <= i; j++) {
if (i != 0 && j != 0)
num = num * (i - j + 1) / j;
printf("%4d", num);
}
printf("</p><p>");
}
}
int main() {
int value = 5;
bin_table(value);
return 0;
}
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