Question

In the shooting competition at Beijing a man could score \[5,4,3,2,1\] or 0 points for each shot. Then the number of ways in which he could score 10 points in seven shot, is
(A) 6538
(B) 6548
(C) 6608
(D) None of these

Answer 1

Hint: According to the given question, firstly assume number of shots be \[{x_i}\] and make the equation to calculate the number of ways that is coefficient of \[{x^{10}}\] in \[{\left( {{x^0} + {x^1} + {x^2} + {x^3} + {x^4} + {x^5}} \right)^7}\]
Then solve the equation by using binomial expansion and hence simplify using the combination by \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] to get the required result.

Complete step-by-step answer:
It is given that in the shooting competition at Beijing a man could score \[5,4,3,2,1\] or 0 points for each shot and we have to find out the number of ways in which he could score 10 points in seven shots.
Let us assume number of shots denoted by \[{x_i}\] and \[{x_i}:\left\{ {0,1,2,3,4,5} \right\}\]
As, it is given that there are 7 seven shots in 10 score points is given by
 \[{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} + {x_7} = 10\]
So, we have to calculate the required number of ways
 \[ \Rightarrow \] Coefficient of \[{x^{10}}\] in \[{\left( {{x^0} + {x^1} + {x^2} + {x^3} + {x^4} + {x^5}} \right)^7}\]
As we know, \[{x^0} = 1\] so we will substitute these values in the above equation. So, we get
 \[ \Rightarrow \] Coefficient of \[{x^{10}}\] in \[{\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5}} \right)^7}\]
Here, we will apply binomial expansion that is given by \[{\left( {\dfrac{{1 - {x^{n + 1}}}}{{1 - x}}} \right)^m}\] . As, \[n = 5\] and \[m = 7\]
On substituting the values we get,
 \[ \Rightarrow \] Coefficient of \[{x^{10}}\] in \[{\left( {\dfrac{{1 - {x^{5 + 1}}}}{{1 - x}}} \right)^7}\]
After simplifying the above equation we get,
 \[ \Rightarrow \] Coefficient of \[{x^{10}}\] in \[{\left( {\dfrac{{1 - {x^6}}}{{1 - x}}} \right)^7}\]
On separating numerator and denominator we get,
 \[ \Rightarrow \] Coefficient of \[{x^{10}}\] in \[{\left( {1 - {x^6}} \right)^7}{\left( {1 - x} \right)^{ - 7}}\]
Opening \[{\left( {1 - {x^6}} \right)^7}\] as:
 \[ \Rightarrow \] Coefficient of \[{x^{10}}\] in \[{\left( {1 - 7{C_1}{x^6} + 7{C_2}{x^{12}} + .....} \right)^7}{\left( {1 - x} \right)^{ - 7}}\]
Rewriting the above equation as,
 \[ \Rightarrow \] \[1 \times \] Coefficient of \[{x^{10}}\] in \[{\left( {1 - x} \right)^{ - 7}} - 7 \times \] coefficient of \[{x^4}\] in \[{\left( {1 - x} \right)^{ - 7}}\]
 \[{ \Rightarrow ^{10 + 7 - 1}}{C_{7 - 1}} - 7{ \times ^{4 + 7 - 1}}{C_{7 - 1}}\]
On simplifying we get,
 \[{ \Rightarrow ^{16}}{C_6} - 7{ \times ^{10}}{C_6}\]
Opening \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
 \[ \Rightarrow \dfrac{{16!}}{{10!6!}} - 7 \times \dfrac{{10!}}{{4!6!}}\]
Now we will open the factorials
 \[ \Rightarrow \dfrac{{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10!}}{{\left( {10!} \right)6 \times 5 \times 4 \times 3 \times 2 \times 1}} - 7 \times \dfrac{{10 \times 9 \times 8 \times 7 \times 6!}}{{4 \times 3 \times 2 \times 1\left( {6!} \right)}}\]
Cancelling \[10!\] and \[6!\]
 \[ \Rightarrow \dfrac{{16 \times 15 \times 14 \times 13 \times 12 \times 11}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}} - 7 \times \dfrac{{10 \times 9 \times 8 \times 7}}{{4 \times 3 \times 2 \times 1}}\]
On simplifying we get,
 \[ \Rightarrow 8008 - 7 \times 210\]
 \[ \Rightarrow 8008 - 1470\]
On subtracting we get,
 \[ \Rightarrow 6538\]
So, the correct answer is “Option A”.

Note: To solve these types of questions you must remember how to expand the binomial expansion and calculate coefficient required in the question. Make the equations according to the given statements and hence put the values in the formulas very carefully.