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Question: Which of the following is/are true? (i)\[{5^6} - {}^5{C_1}{4^6} + {}^5{C_2}{3^6} - {}^5{C_3}{2^6} ...

Which of the following is/are true?
(i)56−5C146+5C236−5C326+5C416=6C2∣! 5 ‾{5^6} - {}^5{C_1}{4^6} + {}^5{C_2}{3^6} - {}^5{C_3}{2^6} + {}^5{C_4}{1^6} = {}^6{C_2}\left| \\!{\underline {\, 5 \,}} \right.
(ii)65−6C155+6C245−6C335+6C425−6C115=0{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_1}{1^5} = 0
(iii)66−6C156+6C246−6C336+6C426−6C516=720{6^6} - {}^6{C_1}{5^6} + {}^6{C_2}{4^6} - {}^6{C_3}{3^6} + {}^6{C_4}{2^6} - {}^6{C_5}{1^6} = 720
(iv)65−6C155+6C245−6C335+6C425−6C515=5C2∣! 6 ‾{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_5}{1^5} = {}^5{C_2}\left| \\!{\underline {\, 6 \,}} \right.

Explanation

Solution

We can separately solve the left hand side and right hand side. You can use the formulas to derive it and show which statements are true or false.

Formula used:
Here, we used the formula nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} , n!=n×(n−1)×....×1n! = n \times \left( {n - 1} \right) \times .... \times 1 .

Complete step-by-step answer:
(i)56−5C146+5C236−5C326+5C416=6C2∣! 5 ‾{5^6} - {}^5{C_1}{4^6} + {}^5{C_2}{3^6} - {}^5{C_3}{2^6} + {}^5{C_4}{1^6} = {}^6{C_2}\left| \\!{\underline {\, 5 \,}} \right.
We will firstly solve all the combinations present in the statement using the combination formula as shown above.
5C1=5!(5−1)!1!^5{C_1} = \dfrac{{5!}}{{\left( {5 - 1} \right)!1!}}
On solving further we get,
5C1=5!4!1!^5{C_1} = \dfrac{{5!}}{{4!1!}}
Solving all the factorial’s factorial formula as shown above.
⇒\Rightarrow 5C1=5×4!4!^5{C_1} = \dfrac{{5 \times 4!}}{{4!}}
Cancelling out 4! From both numerator and denominator:
So, we get 5C1=5^5{C_1} = 5
⇒\Rightarrow 5C2=5!(5−2)!2!^5{C_2} = \dfrac{{5!}}{{\left( {5 - 2} \right)!2!}}
On solving further we get,
5C1=5!3!2!^5{C_1} = \dfrac{{5!}}{{3!2!}}
Solving all the factorials using factorial formula as shown above.
⇒\Rightarrow 5C1=5×4×3!3!×2^5{C_1} = \dfrac{{5 \times 4 \times 3!}}{{3! \times 2}}
Cancelling out 3! From both numerator and denominator:
⇒\Rightarrow 5C1=202^5{C_1} = \dfrac{{20}}{2}
So, we get 5C1=10^5{C_1} = 10
5C3=5!(5−3)!3!^5{C_3} = \dfrac{{5!}}{{\left( {5 - 3} \right)!3!}}
On solving further we get,
⇒\Rightarrow 5C3=5!2!3!^5{C_3} = \dfrac{{5!}}{{2!3!}}
Solving all the factorials using the factorial formula as shown above.
⇒\Rightarrow 5C3=5×4×3!2×3!^5{C_{_3}} = \dfrac{{5 \times 4 \times 3!}}{{2 \times 3!}}
Cancelling out 3! From both numerator and denominator:
⇒\Rightarrow 5C3=202^5{C_3} = \dfrac{{20}}{2}
So, we get 5C3=10^5{C_3} = 10
5C4=5!(5−4)!4!^5{C_4} = \dfrac{{5!}}{{\left( {5 - 4} \right)!4!}}
On solving further we get,
⇒\Rightarrow 5C4=5!1!4!^5{C_4} = \dfrac{{5!}}{{1!4!}}
Solving all the factorial’s factorial formula as shown above.
⇒\Rightarrow 5C4=5×4!4!^5{C_4} = \dfrac{{5 \times 4!}}{{4!}}
Cancelling out 4! From both numerator and denominator:
So, we get 5C4=5^5{C_4} = 5
6C2=6!(6−2)!2!^6{C_2} = \dfrac{{6!}}{{\left( {6 - 2} \right)!2!}}
On solving further we get,
⇒\Rightarrow 6C2=6!4!2!^6{C_2} = \dfrac{{6!}}{{4!2!}}
Solving all the factorial’s factorial formula as shown above.
6C2=6×5×4!4!2!^6{C_2} = \dfrac{{6 \times 5 \times 4!}}{{4!2!}}
Cancelling out 4! From both numerator and denominator:
⇒\Rightarrow 6C2=302^6{C_2} = \dfrac{{30}}{2}
So, we get 6C2=15^6{C_2} = 15
Putting all the combination values in L.H.S ⇒56−5C146+5C236−5C326+5C416 \Rightarrow {5^6} - {}^5{C_1}{4^6} + {}^5{C_2}{3^6} - {}^5{C_3}{2^6} + {}^5{C_4}{1^6}
We get,
56−5×46+10×36−10×26+5×16{5^6} - 5 \times {4^6} + 10 \times {3^6} - 10 \times {2^6} + 5 \times {1^6}
⇒5(55−46+2×36−2×26+16)\Rightarrow 5\left( {{5^5} - {4^6} + 2 \times {3^6} - 2 \times {2^6} + {1^6}} \right)
On simplifying we get,
⇒5(3125−4096+2×729−2×64+1)\Rightarrow 5\left( {3125 - 4096 + 2 \times 729 - 2 \times 64 + 1} \right)
⇒5(3125−4096+1458−128+1)\Rightarrow 5\left( {3125 - 4096 + 1458 - 128 + 1} \right)
⇒5(4584−4224)\Rightarrow 5\left( {4584 - 4224} \right)
⇒5(360)\Rightarrow 5\left( {360} \right)
We get, 56−5C146+5C236−5C326+5C416=1800{5^6} - {}^5{C_1}{4^6} + {}^5{C_2}{3^6} - {}^5{C_3}{2^6} + {}^5{C_4}{1^6} = 1800
Now, we will calculate ⇒6C2∣! 5 ‾\Rightarrow {}^6{C_2}\left| \\!{\underline {\, 5 \,}} \right.
Putting all the values in above, ⇒15×120 \Rightarrow 15 \times 120
We get, 6C2∣! 5 ‾=1800{}^6{C_2}\left| \\!{\underline {\, 5 \,}} \right. = 1800
Hence, L.H.S=R.H.S
Therefore, 56−5C146+5C236−5C326+5C416=6C2∣! 5 ‾{5^6} - {}^5{C_1}{4^6} + {}^5{C_2}{3^6} - {}^5{C_3}{2^6} + {}^5{C_4}{1^6} = {}^6{C_2}\left| \\!{\underline {\, 5 \,}} \right. statement is true.
65−6C155+6C245−6C335+6C425−6C115=0{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_1}{1^5} = 0
Similarly, solve all the combinations as done in part (i)
Solving L.H.S, 65−6C155+6C245−6C335+6C425−6C115{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_1}{1^5}
By putting all the values of combination:
⇒65−6×55+15×45−20×35+15×25−6×15\Rightarrow {6^5} - 6 \times {5^5} + 15 \times {4^5} - 20 \times {3^5} + 15 \times {2^5} - 6 \times {1^5}
⇒7776−6×3125+15×1024−20×243+15×32−6×1\Rightarrow 7776 - 6 \times 3125 + 15 \times 1024 - 20 \times 243 + 15 \times 32 - 6 \times 1
By simplifying:
⇒7776−18750+15360−4860+480−6\Rightarrow 7776 - 18750 + 15360 - 4860 + 480 - 6
⇒23616−23616\Rightarrow 23616 - 23616
We get, 65−6C155+6C245−6C335+6C425−6C115=0{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_1}{1^5} = 0 which is equal to R.H.S.
Therefore, 65−6C155+6C245−6C335+6C425−6C115=0{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_1}{1^5} = 0 statement is true.

66−6C156+6C246−6C336+6C426−6C516=720{6^6} - {}^6{C_1}{5^6} + {}^6{C_2}{4^6} - {}^6{C_3}{3^6} + {}^6{C_4}{2^6} - {}^6{C_5}{1^6} = 720
Similarly, solve all the combinations as done in part (i)
Solving L.H.S, 66−6C156+6C246−6C336+6C426−6C516{6^6} - {}^6{C_1}{5^6} + {}^6{C_2}{4^6} - {}^6{C_3}{3^6} + {}^6{C_4}{2^6} - {}^6{C_5}{1^6}
By putting all the values of combination:
⇒66−6×56+15×46−20×36+15×26−6×16\Rightarrow {6^6} - 6 \times {5^6} + 15 \times {4^6} - 20 \times {3^6} + 15 \times {2^6} - 6 \times {1^6}
⇒46656−6×15625+15×4096−20×729+15×64−6×1\Rightarrow 46656 - 6 \times 15625 + 15 \times 4096 - 20 \times 729 + 15 \times 64 - 6 \times 1
By simplifying:
⇒46656−93750+61440−14580+960−6\Rightarrow 46656 - 93750 + 61440 - 14580 + 960 - 6
⇒109056−108336\Rightarrow 109056 - 108336
We get, 66−6C156+6C246−6C336+6C426−6C516=720{6^6} - {}^6{C_1}{5^6} + {}^6{C_2}{4^6} - {}^6{C_3}{3^6} + {}^6{C_4}{2^6} - {}^6{C_5}{1^6} = 720 which is equal to R.H.S.
Therefore, 66−6C156+6C246−6C336+6C426−6C516=720{6^6} - {}^6{C_1}{5^6} + {}^6{C_2}{4^6} - {}^6{C_3}{3^6} + {}^6{C_4}{2^6} - {}^6{C_5}{1^6} = 720 statement is true.

(iv)65−6C155+6C245−6C335+6C425−6C515=5C2∣! 6 ‾{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_5}{1^5} = {}^5{C_2}\left| \\!{\underline {\, 6 \,}} \right.
Similarly, solve all the combinations as done in part (i)
Solving L.H.S, 65−6C155+6C245−6C335+6C425−6C515{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_5}{1^5}
By putting all the values of combination:
⇒65−6×55+15×45−20×35+15×25−6×15\Rightarrow {6^5} - 6 \times {5^5} + 15 \times {4^5} - 20 \times {3^5} + 15 \times {2^5} - 6 \times {1^5}
⇒7776−6×3125+15×1024−20×243+15×32−6×1\Rightarrow 7776 - 6 \times 3125 + 15 \times 1024 - 20 \times 243 + 15 \times 32 - 6 \times 1
By simplifying:
⇒7776−18750+15360−4860+480−6\Rightarrow 7776 - 18750 + 15360 - 4860 + 480 - 6
⇒23616−23616\Rightarrow 23616 - 23616
We get, 65−6C155+6C245−6C335+6C425−6C515=0{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_5}{1^5} = 0
On solving R.H.S 5C2∣! 6 ‾{}^5{C_2}\left| \\!{\underline {\, 6 \,}} \right. we get,
⇒10×6!\Rightarrow 10 \times 6!
⇒10×720\Rightarrow 10 \times 720
⇒7200\Rightarrow 7200
Hence, L.H.S ≠\neR.H.S
Therefore, 65−6C155+6C245−6C335+6C425−6C515=5C2∣! 6 ‾{6^5} - {}^6{C_1}{5^5} + {}^6{C_2}{4^5} - {}^6{C_3}{3^5} + {}^6{C_4}{2^5} - {}^6{C_5}{1^5} = {}^5{C_2}\left| \\!{\underline {\, 6 \,}} \right. statement is false.

Note: These types of questions are done with the help of combination and factorial formulas. Do the calculation very carefully and make the calculations as simple as you can. Questions like these can be lengthy but you have to follow the same steps most of the cases.