C3 ⇒50C4+55C3+54C3+53C3+52C3+51C3+50C3 Arrange those this way ⇒50C4+50C3+51C3+52C3+53C3+54C3+55C3 We know this formula [nCr+nCr−1=n+1Cr] which is used to solve this problem. ⇒51C4+51C3+52C3+53C3+54C3+55C3 ⇒52C4+52C3+53C3+54C3+55C3 ⇒53C4+53C3+54C3+55C3 ⇒54C4+54C3+55C3 ⇒55C4+55C3 ⇒56C4