The total combinations of picking $3$ balloons from a packet of $25$ balloons are
Answer
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Hint: A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order. The formula of the combination is $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ . We find the combination by using this formula and we get the required result.
Complete step-by-step answer:
Given that we find the combinations of picking $3$ balloons from a packet of $25$ balloons .
Therefore we find the combination $^{25}{C_3}$
We use the formula of the combination i.e., $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ , we get
$^{23}{C_3} = \dfrac{{25!}}{{3!(25 - 3)!}}$
Expanding the above factorial and we get
$ = \dfrac{{25 \times 24 \times 23 \times 22!}}{{3 \times 2 \times 1 \times 22!}}$
Now we omitting the same term and we get
$ = \dfrac{{25 \times 24 \times 23}}{{3 \times 2}}$
Simplifying the denominator and we get
$ = \dfrac{{25 \times 24 \times 23}}{6}$
Now we divide $24$ by $6$ , therefore we get $\dfrac{{24}}{6} = 4$
Put this in above function and we get
$ = 25 \times 23 \times 4$
Multiplying $25$ and $4$ , we get
$ = 100 \times 23$
Simplifying and we get
$ = 2300$
So, the correct answer is “2300”.
Note: Factorial is the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point. Thus , factorial $n$ is written as $n!$ and defined as $n! = n \times (n - 1) \times (n - 2) \times ...... \times 3 \times 2 \times 1$ .
For this type of problem, we use only the combination formula to get the answer.
Complete step-by-step answer:
Given that we find the combinations of picking $3$ balloons from a packet of $25$ balloons .
Therefore we find the combination $^{25}{C_3}$
We use the formula of the combination i.e., $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ , we get
$^{23}{C_3} = \dfrac{{25!}}{{3!(25 - 3)!}}$
Expanding the above factorial and we get
$ = \dfrac{{25 \times 24 \times 23 \times 22!}}{{3 \times 2 \times 1 \times 22!}}$
Now we omitting the same term and we get
$ = \dfrac{{25 \times 24 \times 23}}{{3 \times 2}}$
Simplifying the denominator and we get
$ = \dfrac{{25 \times 24 \times 23}}{6}$
Now we divide $24$ by $6$ , therefore we get $\dfrac{{24}}{6} = 4$
Put this in above function and we get
$ = 25 \times 23 \times 4$
Multiplying $25$ and $4$ , we get
$ = 100 \times 23$
Simplifying and we get
$ = 2300$
So, the correct answer is “2300”.
Note: Factorial is the product of all positive integers less than or equal to a given positive integer and denoted by that integer and an exclamation point. Thus , factorial $n$ is written as $n!$ and defined as $n! = n \times (n - 1) \times (n - 2) \times ...... \times 3 \times 2 \times 1$ .
For this type of problem, we use only the combination formula to get the answer.
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