Solveeit Logo
Login

Question

Question: Verify that \({}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}={}^{11}{{C}_{4}}\) ....

Verify that 10C4+10C3=11C4{}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}={}^{11}{{C}_{4}} .

Explanation

Solution

We first need to write down the formula for nCr{}^{n}{{C}_{r}} which is nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} . Then, we need to put values n=10,10,11n=10,10,11 and r=4,3,4r=4,3,4 respectively to find the values of 10C4,10C3,11C4{}^{10}{{C}_{4}},{}^{10}{{C}_{3}},{}^{11}{{C}_{4}} . Having done so, we then need to check the value of the LHS and see if it equals the RHS. If yes, then our expression is verified.

Complete step-by-step solution:
The expression that we are given in this problem is,
10C4+10C3=11C4{}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}={}^{11}{{C}_{4}}
We can prove the expression by equating the LHS and the RHS of the equation. But, before that, we need to simplify each of the sides to a natural number each and then show that the two natural numbers are one and the same.
We know that the formula for nCr{}^{n}{{C}_{r}} is nCr=n!r!(n−r)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} . Now, we need to put some values in this formula one by one to find the values of 10C4,10C3,11C4{}^{10}{{C}_{4}},{}^{10}{{C}_{3}},{}^{11}{{C}_{4}} . At first, putting n=10,r=3n=10,r=3 , we get,
⇒10C3=10!3!(10−3)! ⇒10C3=10!3!×7!=10×9×83×2×1=120 \begin{aligned} & \Rightarrow {}^{10}{{C}_{3}}=\dfrac{10!}{3!\left( 10-3 \right)!} \\\ & \Rightarrow {}^{10}{{C}_{3}}=\dfrac{10!}{3!\times 7!}=\dfrac{10\times 9\times 8}{3\times 2\times 1}=120 \\\ \end{aligned}
Then, putting n=10,r=4n=10,r=4 , we get,
⇒10C4=10!4!(10−4)! ⇒10C4=10!4!×6!=10×9×8×74×3×2×1=210 \begin{aligned} & \Rightarrow {}^{10}{{C}_{4}}=\dfrac{10!}{4!\left( 10-4 \right)!} \\\ & \Rightarrow {}^{10}{{C}_{4}}=\dfrac{10!}{4!\times 6!}=\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}=210 \\\ \end{aligned}
The, putting n=11,r=4n=11,r=4 , we get,
⇒11C4=11!4!(11−4)! ⇒11C4=11!4!×7!=11×10×9×84×3×2×1=330 \begin{aligned} & \Rightarrow {}^{11}{{C}_{4}}=\dfrac{11!}{4!\left( 11-4 \right)!} \\\ & \Rightarrow {}^{11}{{C}_{4}}=\dfrac{11!}{4!\times 7!}=\dfrac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}=330 \\\ \end{aligned}
The LHS becomes,
⇒10C4+10C3=120+210=330\Rightarrow {}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}=120+210=330
And, the RHS becomes,
⇒11C4=330\Rightarrow {}^{11}{{C}_{4}}=330
This means that the left hand side has become equal to the right hand side. This means that the expression is verified.
Thus, we can conclude that 10C4+10C3=11C4{}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}={}^{11}{{C}_{4}} .

Note: We can also solve the problem in another way. There is a property of combinations which goes as nCr+nCr−1=n+1Cr{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} . We just simply need to check whether this relation is valid for this problem or not. By simple comparison, we can see that in our problem, n=10,r=4n=10,r=4 . So, the property is applicable for our problem and thus the expression is verified.