Answer
Verified
39.3k+ views
Hint: Form the cases based on the given condition that out of 21 matches played by the team 19 matches are won. After making the cases, find the number of possible ways that event or case can occur. And in the end add all the values obtained in each of the cases.
Complete step-by-step solution
Let us begin with the fact that out of 21 football matches played by the team, 19 matches have been won by the team. Thus, we can observe that there can be three cases to find the number of different forecasts that contain 19 wins.
Those cases can be listed as:
1. There are 19 wins and 2 losses.
2. There are 19 wins, 1 loss and 1 draw.
3. There are 19 wins and 2 draws.
Let us consider these cases one by one and find out their possible outcomes.
Case 1:- There are 19 wins and 2 losses.
The number of possible ways is given by dividing 21! by the possibility 19 wins, that is 19! multiplied by the possibility of 2 losses, that is 2!
$
\Rightarrow N = \dfrac{{21!}}{{19! \times 2!}} \\
= \dfrac{{21 \times 20}}{2} \\
= 210 \\
$
Case 2:- There are 19 wins, 1 loses and 1 draw.
The number of possible ways is given by dividing 21! by the possibility 19 wins, that is 19! multiplied by the possibility of 1 lose, that is 1! and the possibility of 1 draw, that is, 1!.
$
\Rightarrow N = \dfrac{{21!}}{{19! \times 1! \times 1!}} \\
= \dfrac{{21 \times 20}}{1} \\
= 420 \\
$
Case 3:- There are 19 wins and 2 draws.
The number of possible ways is given by dividing 21! by the possibility 19 wins, that is 19! multiplied by the possibility of 2 draws, that is 2!
$
\Rightarrow N = \dfrac{{21!}}{{19! \times 2!}} \\
= \dfrac{{21 \times 20}}{2} \\
= 210 \\
$
Thus, now we add all the possible ways obtained in all the 3 cases to get the no. of different forecasts that can contain 19 wins.
$
\Rightarrow {\text{Total}} = 210 + 420 + 210 \\
= 840 \\
$
Hence, option (C) is the correct option.
Note: Listing down all the possible ways is a cumbersome and time taking task, thus we use the method of permutation or fundamentals of counting principles to solve questions like these. Make sure to analyze everything properly to not skip any case. Also, make your concept very clear about where to apply the repetition method and where not to. Also, do not mix combinations with permutation. Do not make calculation errors.
Complete step-by-step solution
Let us begin with the fact that out of 21 football matches played by the team, 19 matches have been won by the team. Thus, we can observe that there can be three cases to find the number of different forecasts that contain 19 wins.
Those cases can be listed as:
1. There are 19 wins and 2 losses.
2. There are 19 wins, 1 loss and 1 draw.
3. There are 19 wins and 2 draws.
Let us consider these cases one by one and find out their possible outcomes.
Case 1:- There are 19 wins and 2 losses.
The number of possible ways is given by dividing 21! by the possibility 19 wins, that is 19! multiplied by the possibility of 2 losses, that is 2!
$
\Rightarrow N = \dfrac{{21!}}{{19! \times 2!}} \\
= \dfrac{{21 \times 20}}{2} \\
= 210 \\
$
Case 2:- There are 19 wins, 1 loses and 1 draw.
The number of possible ways is given by dividing 21! by the possibility 19 wins, that is 19! multiplied by the possibility of 1 lose, that is 1! and the possibility of 1 draw, that is, 1!.
$
\Rightarrow N = \dfrac{{21!}}{{19! \times 1! \times 1!}} \\
= \dfrac{{21 \times 20}}{1} \\
= 420 \\
$
Case 3:- There are 19 wins and 2 draws.
The number of possible ways is given by dividing 21! by the possibility 19 wins, that is 19! multiplied by the possibility of 2 draws, that is 2!
$
\Rightarrow N = \dfrac{{21!}}{{19! \times 2!}} \\
= \dfrac{{21 \times 20}}{2} \\
= 210 \\
$
Thus, now we add all the possible ways obtained in all the 3 cases to get the no. of different forecasts that can contain 19 wins.
$
\Rightarrow {\text{Total}} = 210 + 420 + 210 \\
= 840 \\
$
Hence, option (C) is the correct option.
Note: Listing down all the possible ways is a cumbersome and time taking task, thus we use the method of permutation or fundamentals of counting principles to solve questions like these. Make sure to analyze everything properly to not skip any case. Also, make your concept very clear about where to apply the repetition method and where not to. Also, do not mix combinations with permutation. Do not make calculation errors.
Recently Updated Pages
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
The number of ways in which 5 boys and 3 girls can-class-12-maths-JEE_Main
Find dfracddxleft left sin x rightlog x right A left class 12 maths JEE_Main
Distance of the point x1y1z1from the line fracx x2l class 12 maths JEE_Main
In a box containing 100 eggs 10 eggs are rotten What class 12 maths JEE_Main
dfracddxex + 3log x A ex cdot x2x + 3 B ex cdot xx class 12 maths JEE_Main
Other Pages
Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main
when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
448 litres of methane at NTP corresponds to A 12times class 11 chemistry JEE_Main
A convex lens is dipped in a liquid whose refractive class 12 physics JEE_Main