`=>` Let us now assume that there is a group of `3` lawn tennis players `color(blue)(X, Y, Z.)` A team consisting of `2` players is to be formed.
`color(blue)("In how many ways can we do so ?")`
Here, `color(blue)("order is not important.")` In fact, there are only `3` possible ways in which the team could be constructed.
These are `color(blue)(XY, YZ)` and `color(blue)(ZX)` (Fig 1. )
Here, each selection is called a combination of `3` different objects taken `2` at a time. In a combination, the order is not important.
Some more illustrations where order is not important
Twelve persons meet in a room and each shakes hand with all the others. How do we determine the number of hand shakes. X shaking hands with Y and Y with X will not be two different hand shakes. Here, order is not important. There will be as many hand shakes as there are combinations of 12 different things taken 2 at a time.
Seven points lie on a circle. How many chords can be drawn by joining thesepoints pairwise? There will be as many chords as there are combinations of 7 different things taken 2 at a time.
Now, we obtain the formula for finding the number of combinations of n different objects taken r at a time, denoted by `color(red)(text()^nC_r).`
`=>` `color(green)("Suppose we have 4 different objects A, B, C and D. Taking 2 at a time")`,
if we have to make combinations, these will be `AB, AC, AD, BC, BD, CD.`
Here, `AB` and `BA` are the same combination as order does not alter the combination.
This is why we have not included `BA, CA, DA, CB, DB` and `DC` in this list.
There are as many as `6` combinations of `4` different objects taken `2` at a time, i.e., `text()^4C_2 = 6.`
Corresponding to each combination in the list, we can arrive at `2!` permutations as `2` objects in each combination can be rearranged in `2!` ways. Hence, the number of permutations `= text()^4C_2 × 2!.`
On the other hand, the number of permutations of `4` different things taken `2` at a time `= text()^4P_2.`
Therefore `text()^4P_2 = text()^4C_2 × 2!` or `(4!)/((4-2)! 2!)= text()^4C_2 `
`\color{purple}ul(✓✓) \color{purple} " DEFINITION ALERT" `
`color(red)(★ ul"Combinations")`
Combination is selection of one or more things out of n things which may be alike or different taken some oral! at a time.
it is denoted by `color(red)(text()^nC_r).`
`=>` Let us now assume that there is a group of `3` lawn tennis players `color(blue)(X, Y, Z.)` A team consisting of `2` players is to be formed.
`color(blue)("In how many ways can we do so ?")`
Here, `color(blue)("order is not important.")` In fact, there are only `3` possible ways in which the team could be constructed.
These are `color(blue)(XY, YZ)` and `color(blue)(ZX)` (Fig 1. )
Here, each selection is called a combination of `3` different objects taken `2` at a time. In a combination, the order is not important.
Some more illustrations where order is not important
Twelve persons meet in a room and each shakes hand with all the others. How do we determine the number of hand shakes. X shaking hands with Y and Y with X will not be two different hand shakes. Here, order is not important. There will be as many hand shakes as there are combinations of 12 different things taken 2 at a time.
Seven points lie on a circle. How many chords can be drawn by joining thesepoints pairwise? There will be as many chords as there are combinations of 7 different things taken 2 at a time.
Now, we obtain the formula for finding the number of combinations of n different objects taken r at a time, denoted by `color(red)(text()^nC_r).`
`=>` `color(green)("Suppose we have 4 different objects A, B, C and D. Taking 2 at a time")`,
if we have to make combinations, these will be `AB, AC, AD, BC, BD, CD.`
Here, `AB` and `BA` are the same combination as order does not alter the combination.
This is why we have not included `BA, CA, DA, CB, DB` and `DC` in this list.
There are as many as `6` combinations of `4` different objects taken `2` at a time, i.e., `text()^4C_2 = 6.`
Corresponding to each combination in the list, we can arrive at `2!` permutations as `2` objects in each combination can be rearranged in `2!` ways. Hence, the number of permutations `= text()^4C_2 × 2!.`
On the other hand, the number of permutations of `4` different things taken `2` at a time `= text()^4P_2.`
Therefore `text()^4P_2 = text()^4C_2 × 2!` or `(4!)/((4-2)! 2!)= text()^4C_2 `
`\color{purple}ul(✓✓) \color{purple} " DEFINITION ALERT" `
`color(red)(★ ul"Combinations")`
Combination is selection of one or more things out of n things which may be alike or different taken some oral! at a time.
it is denoted by `color(red)(text()^nC_r).`