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Question: How do you evaluate \({}^{11}{{C}_{7}}\) ?...

How do you evaluate 11C7{}^{11}{{C}_{7}} ?

Explanation

Solution

To evaluate the given question 11C7{}^{11}{{C}_{7}} , we will use a formula that is:
⇒nCr=n!r!(n−r)!\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} , Where, 0≤r≤n0\le r\le n
And factorial nn denoted as n!n! represents the multiplication of all the numbers up to nn as:
⇒n!=1×2×3×...×(n−1)×n\Rightarrow n!=1\times 2\times 3\times ...\times \left( n-1 \right)\times n
Similarly,
⇒r!=1×2×3×...×(r−1)×r\Rightarrow r!=1\times 2\times 3\times ...\times \left( r-1 \right)\times r

Complete step by step solution:
Since, we have the given question as 11C7{}^{11}{{C}_{7}} in the form of nCr{}^{n}{{C}_{r}} . After comparison of 11C7{}^{11}{{C}_{7}} and nCr{}^{n}{{C}_{r}}, we will the value for nn and rr as:
⇒n=11\Rightarrow n=11
And
⇒r=7\Rightarrow r=7
Here, we use the formula for getting the value of given question as:
⇒nCr=n!r!(n−r)!\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
Now, we will apply the value1111 and 77 in the place of nn and rr respectively as:
⇒11C7=11!7!(11−7)!\Rightarrow {}^{11}{{C}_{7}}=\dfrac{11!}{7!\left( 11-7 \right)!}
Since, we replaced the value of nn and rr in the formula. Then we solve the bracketed numbers by using subtraction. Thus, we will get 44 after subtracting 77 from 1111 as:
⇒11−7=4\Rightarrow 11-7=4
So, we will use this value in the formula as:
⇒11C7=11!7!×4!\Rightarrow {}^{11}{{C}_{7}}=\dfrac{11!}{7!\times 4!}
Now, we will expand the all the factorial numbers respectively as:
⇒11!=1×2×3×4×5×6×7×8×9×10×11\Rightarrow 11!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10\times 11
Here we can write 7!7! for the multiplication of first 77 numbers that will help to cancel out 7!7! in the formula as:
⇒11!=7!×8×9×10×11\Rightarrow 11!=7!\times 8\times 9\times 10\times 11
Further we will do multiplication for next 44 numbers and will get 79207920 after multiplication. So we can write the above expression as:
⇒11!=7!×7920\Rightarrow 11!=7!\times 7920
Similarly, we will do the expression and multiplication for 4!4! as:
⇒4!=1×2×3×4\Rightarrow 4!=1\times 2\times 3\times 4
⇒4!=24\Rightarrow 4!=24
Now, we will apply the values of all factorial except 7!7! because we will eliminate 7!7! as:
⇒11C7=7!×79207!×24\Rightarrow {}^{11}{{C}_{7}}=\dfrac{7!\times 7920}{7!\times 24}
Here, we wrote 11!11! into the form of multiple of 7!7! in the above formula so that we can cancel out 7!7! as:
⇒11C7=792024\Rightarrow {}^{11}{{C}_{7}}=\dfrac{7920}{24}
Now, we will divide 79207920 from 2424 that will divide the number 79207920 completely and will get quotient 330330 as:
⇒11C7=330\Rightarrow {}^{11}{{C}_{7}}=330
Hence, the solution for the 11C7{}^{11}{{C}_{7}} is 330330 .

Note: Since, we read the combination with permutation in the chapter named as permutation and combination, we need to completely understand the difference between these two. Permutation is the method that helps us to know the possible numbers of way for arranging the data in a sequence, while combination helps us to know the possible numbers of way of section of data and both have different formula that we need to learn that are
For permutation:
⇒nPr=n!(n−r)!\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}
And for combination:
⇒nCr=n!r!(n−r)!\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}