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Question: In how many ways can a girl and a boy be selected from a group of \(15\) boys and \(10\) girls? A....

In how many ways can a girl and a boy be selected from a group of 1515 boys and 1010 girls?
A. 15×1015 \times 10
B. 15+1015 + 10
C. 25P2^{25}{P_2}
D. 25C2^{25}{C_2}

Explanation

Solution

As we can see that the above question is related to Permutation and Combination. So in this question we will use the formula for the number of ways for selecting rr things from nn group of people. The formula of combination is given as; nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}. So we will use this formula to solve the given problem.

Complete step by step answer:
In the above question we have a total number of boys 1515.
And we have to select one boy from a group of 1515 people. Here we have :
n=15,r=1n = 15,r = 1
Now by putting the value in the formula we can write:
n!r!(n−r)!=15!1!(15−1)!\dfrac{{n!}}{{r!\left( {n - r} \right)!}} = \dfrac{{15!}}{{1!(15 - 1)!}}
On simplifying it gives us :
15!1!×14!\dfrac{{15!}}{{1! \times 14!}}
Now we can break the factorial values and it gives the value:
15×14!14!=15\dfrac{{15 \times 14!}}{{14!}} = 15
Again we have a total number of girls given 1010.

And we have to select one girl from a group of 1010 girls. Here we have :
n=10,r=1n = 10,r = 1
Similarly as above by putting the value in the formula we can write:
n!r!(n−r)!=10!1!(10−1)!\dfrac{{n!}}{{r!\left( {n - r} \right)!}} = \dfrac{{10!}}{{1!(10 - 1)!}}
On simplifying it gives us :
10!1!×9!\dfrac{{10!}}{{1! \times 9!}}
Now we can break the factorial values and it gives the value:
10×9!9!=10\dfrac{{10 \times 9!}}{{9!}} = 10
Now we will multiply both the values to get the required answer:
15C1×10C1^{15}{C_1}{ \times ^{10}}{C_1}
By putting the values of both the combination, we have:
15×1015 \times 10

Hence the correct option is A.

Note: We should note that for selection purposes we use combination and for arranging the values we use the permutation. We should know that if the order does not matter then we use the combination formula as in the above question, but if the order does matter then we use the permutation formula. The value of permutation is denoted by
nPr^n{P_r}. The formula of permutation is n!(n−r)!\dfrac{{n!}}{{\left( {n - r} \right)!}}.