Question

01. Do
you
find that all living beings need the
same kind of food ? ?
​

Answer 1

Answer:

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

★ Concept :

Here the concept of Combinations has been used. Firstly here we shall calculate all the cases for each combination. Then we shall add them. After adding them, we will get the total number of combinations.

Let's do it !!

______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{^{n}C_{r}\;=\;\dfrac{n!}{r!(n\:-\:r)!}}}}$

______________________________________

★ Solution :-

Given,

» Number of persons to be selected = 3

» Total number of males = 5

» Total number of females = 4

We know that,

$\;\bf{\mapsto\;\;Possible\;\:Combinations\;=\;^{n}C_{r}}$

And,

$\;\sf{\rightarrow\;\;^{n}C_{r}\;=\;\dfrac{n!}{r!(n\:-\:r)!}}$

n is the total number of items

r is the number of items selected

C is the combination for those number of selected items

So,

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{n!}{r!(n\:-\:r)!}}$

We shall use this formula to solve every part.

• Total combinations of choosing atleast 1 male ::

Let's find each case seperately where we will have atleast 1 male and then add all the cases.

>> 3 male and 0 female :

Here 3 male out of 5 males are selected and 0 female out of 4 females are selected which shows that atleast 1 male is selected.

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{n!}{r!(n\:-\:r)!}\:\times\:\dfrac{n!}{r!(n\:-\:r)!}}$

Here we have combined combinations for both male and female.

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{5!}{3!(5\:-\:3)!}\:\times\:\dfrac{4!}{0!(4\:-\:0)!}}$

Since 0! = 1

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{5!}{3!2!}\:\times\:\dfrac{4!}{4!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{120}{12}\:\times\:1}$

[5! = 5×4×3×2×1 = 120, 3! = 3×2×1 = 6, 2! = 2×1 = 2]

$\;\bf{\mapsto\;\;Possible\;\:Combinations\;=\;12}$

>> 2 male and 1 female ::

Again using same process, we get

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{5!}{3!(5\:-\:2)!}\:\times\:\dfrac{4!}{1!(4\:-\:1)!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{5!}{2!3!}\:\times\:\dfrac{4!}{1!(4\:-\:1)!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{120}{12}\:\times\:\dfrac{24}{3!}}$

[4! = 4×3×2×1 = 24, 1! = 1]

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{120}{12}\:\times\:\dfrac{24}{6}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;10\:\times\:4}$

$\;\bf{\mapsto\;\;Possible\;\:Combinations\;=\;40}$

>> 1 male and 2 female ::

Again using the same process, we get

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{5!}{1!(5\:-\:1)!}\:\times\:\dfrac{4!}{2!(4\:-\:2)!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{5!}{1!4!}\:\times\:\dfrac{4!}{2!2!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;5\:\times\:6}$

$\;\bf{\mapsto\;\;Possible\;\:Combinations\;=\;30}$

Now we can't take the fourth condition as 0 male and 4 females since the condition of qúestion that is atleast of 1 male will not satisfy.

Now let's add all the possible combinations to find the total number of Combination.

→ Total number of combinations = 10 + 40 + 30

→ Total number of combinations = 80

$\;\underline{\boxed{\tt{Number\;\:of\;\:Combinations\;=\;\bf{\purple{80}}}}}$

-----------------------------------------------------------

A.) Choosing 1 male in advance

Here the condition is that 1 male is already choosen. This means now total males left to be selected is 5 - 1 = 4. Also now only 2 places are left out of 3.

Let's find the Combinations of all the cases.

>> 2 males and 0 females

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{4!}{2!(4\:-\:2)!}\:\times\:\dfrac{4!}{0!(4\:-\:0)!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{4!}{2!2!}\:\times\:\dfrac{4!}{0!4!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{24}{4}\:\times\:1}$

$\;\bf{\mapsto\;\;Possible\;\:Combinations\;=\;6}$

>> 1 male and 1 female

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{4!}{1!(4\:-\:1)!}\:\times\:\dfrac{4!}{1!(4\:-\:1)!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;\dfrac{4!}{1!3!}\:\times\:\dfrac{4!}{1!3!}}$

$\;\sf{\mapsto\;\;Possible\;\:Combinations\;=\;4\:\times\:4}$

[tex]\;\bf{\mapsto\;\;