We know the Pascal’s Identity very well, i.e. ncr = n-1cr + n-1cr-1
A curious reader might have observed that Pascal’s Identity is instrumental in establishing recursive relation in solving binomial coefficients. It is quite easy to prove the above identity using simple algebra. Here I’m trying to explain it’s practical significance.
Recap from counting techniques, ncr means selecting r elements from n elements. Let us pick a special element k from these n elements, we left with (n – 1) elements.
We can group these r elements selection ncr into two categories,
1) group that contains the element k.
2) group that does not contain the element k.
Consider first group, the special element k is in all r selections. Since k is part of r elements, we need to choose (r – 1) elements from remaining (n – 1) elements, there are n-1cr-1 ways.
Consider second group, the special element k is not there in all r selections, i.e. we will have to select all the r elements from available (n – 1) elements (as we must exclude element k from n). This can be done in n-1cr ways.
Now it is evident that sum of these two is selecting r elements from n elements.
There can be many ways to prove the above fact. There might be many applications of Pascal’s Identity. Please share your knowledge.
— Venki. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details
- Cassini’s Identity
- Program for Identity Matrix
- Euler's Four Square Identity
- Brahmagupta Fibonacci Identity
- Proizvolov's Identity
- Factorial of Large numbers using Logarithmic identity
- Minimum area of square holding two identical rectangles
- Number of ways to obtain each numbers in range [1, b+c] by adding any two numbers in range [a, b] and [b, c]
- Count of subarrays of size K which is a permutation of numbers from 1 to K
- Print all permutations of a number N greater than itself
- Create an array such that XOR of subarrays of length K is X
- Find the K closest points to origin using Priority Queue
- Split the number N by maximizing the count of subparts divisible by K
- Count subarrays with sum equal to its XOR value