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Question: The value of \({}^6{C_4}\) is- A) \(6\) B) \(9\) C) \(15\) D) \(240\)...

The value of 6C4{}^6{C_4} is-
A) 66
B) 99
C) 1515
D) 240240

Explanation

Solution

The number of ways to select r things from n things is represented as nCr{}^n{C_r} and also we know that the formula of combination is-
nCr=n!r!n−r!{}^n{C_r} = \dfrac{{n!}}{{r!n - r!}} where n= total number of things and r is the number of things to be selected and also n!=n(n−1)...3,2,1n! = n\left( {n - 1} \right)...3, 2, 1. Use this formula to solve the given question.

Complete step by step solution:
We have to find the value of 6C4{}^6{C_4}.
We know that the formula of combination is-
⇒nCr=n!r!n−r!\Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}-- (i)
Where n= total number of things and r is the number of things to be selected.
So here n=66and r=44
So on putting these values in eq. (i), we get-
⇒6C4=6!4!6−4!\Rightarrow {}^6{C_4} = \dfrac{{6!}}{{4!6 - 4!}}
On performing subtraction in the denominator, we get-
⇒6C4=6!4!2!\Rightarrow {}^6{C_4} = \dfrac{{6!}}{{4!2!}}
Now we know that n!=n(n−1)...3,2,1n! = n\left( {n - 1} \right)...3,2,1
So we can write-
⇒6C4=6×(6−1)×(6−2)×(6−3)×(6−4)×(6−5)4×(4−1)×(4−2)×(4−3)×2×(2−1)\Rightarrow {}^6{C_4} = \dfrac{{6 \times \left( {6 - 1} \right) \times \left( {6 - 2} \right) \times \left( {6 - 3} \right) \times \left( {6 - 4} \right) \times \left( {6 - 5} \right)}}{{4 \times \left( {4 - 1} \right) \times \left( {4 - 2} \right) \times \left( {4 - 3} \right) \times 2 \times \left( {2 - 1} \right)}}
On solving, we get-
⇒6C4=6×5×4×3×2×14×3×2×1×2×1\Rightarrow {}^6{C_4} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}}
On cancelling the same terms of numerator and denominator, we get-
⇒6C4=6×52×1\Rightarrow {}^6{C_4} = \dfrac{{6 \times 5}}{{2 \times 1}}
On multiplication, we get-
⇒6C4=302\Rightarrow {}^6{C_4} = \dfrac{{30}}{2}
On dividing the numerator by denominator, we get-
⇒6C4=15\Rightarrow {}^6{C_4} = 15

Hence the correct answer is option C.

Note:
Students may get confused between the formula of combination and permutation as both look almost the same but there is a difference between the two formulae. The combination is only concerned with selection not order while in permutation order is important. A permutation is concerned with the arrangement of things and it is given as-
⇒nPr=n!n−r!\Rightarrow {}^n{P_r} = \dfrac{{n!}}{{n - r!}}
Where n is the total number of things and r is the number of things to be selected. So we can also write the formula of combination as-
⇒nCr=nPrr!\Rightarrow {}^n{C_r} = \dfrac{{{}^n{P_r}}}{{r!}}
So we can also use this formula to solve the given question.