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Question: How do you evaluate \[^{11}{C_7}\]?...

How do you evaluate 11C7^{11}{C_7}?

Explanation

Solution

We use the formula of combination and substitute the value of n and r as given in the question. Calculate the value using the formula of factorial. Write expanded factorial in the numerator and cancel the same terms from numerator and denominator.

  • Combination is given by nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}, where n is the total number of available objects and r is the number of objects we have to choose.
  • Factorial terms open up as n!=n(n−1)!n! = n(n - 1)!

Complete step-by-step answer:
We have to find the value of 11C7^{11}{C_7}
We compare the given combination function to general combination function i.e. nCr^n{C_r}
So, we get the value of n as 11 and r as 7
Now we substitute the value of n and r in the general formula of combination i.e. nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
⇒11C7=11!(11−7)!7!{ \Rightarrow ^{11}}{C_7} = \dfrac{{11!}}{{(11 - 7)!7!}}
Calculate the difference in the numerator of right hand side of the equation
⇒11C7=11!4!7!{ \Rightarrow ^{11}}{C_7} = \dfrac{{11!}}{{4!7!}} … (1)
Now we expand the numerator using the formula of factorial such that we can cancel out the highest factorial in the denominator.
Here, we see the highest factorial in the denominator is 7, so we expand the factorial in the numerator i.e. 11 till we get factorial of 7.
We have the formula n!=n(n−1)!n! = n(n - 1)!
We can write 11!=11(11−1)(11−2)(11−3)(11−4)!11! = 11(11 - 1)(11 - 2)(11 - 3)(11 - 4)!
i.e. 11!=11×10×9×8×7!11! = 11 \times 10 \times 9 \times 8 \times 7!
Substitute this value of factorial of 11 in equation (1)
⇒11C7=11×10×9×8×7!4!7!{ \Rightarrow ^{11}}{C_7} = \dfrac{{11 \times 10 \times 9 \times 8 \times 7!}}{{4!7!}}
Cancel same factors from numerator and denominator
⇒11C7=11×10×9×84!{ \Rightarrow ^{11}}{C_7} = \dfrac{{11 \times 10 \times 9 \times 8}}{{4!}}
Expand the factorial in the denominator as 4!=4×3×2×14! = 4 \times 3 \times 2 \times 1
⇒11C7=11×10×9×84×3×2×1{ \Rightarrow ^{11}}{C_7} = \dfrac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2 \times 1}}
Cancel same factors from numerator and denominator
⇒11C7=11×10×3{ \Rightarrow ^{11}}{C_7} = 11 \times 10 \times 3
Calculate the product
⇒11C7=330{ \Rightarrow ^{11}}{C_7} = 330

∴\therefore The value of 11C7^{11}{C_7} is 330.

Note:
Many students make the mistake of opening the factorial wrong as they don’t move in increasing order of subtraction from the number and end up with the wrong answer. Keep in mind we subtract 1, 2, 3, and so on from the given number and multiply the terms.