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'sdo it !!
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★FormulaUsed :-
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★Solution:-
Given,
» Number of persons to be selected = 3
»Total number of males = 5
» Total number of females = 4
We know that,
And,
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,
We shall use this formula to solve every part.
• Totalcombinationsof choosingatleast1male ::
Let's find each case seperately where we will have atleast 1 male and then add all the cases.
>>3male and 0female :
Here 3male out of 5males are selected and 0female out of 4females are selected which shows that atleast 1male is selected.
Here we have combined combinations for both male and female.
In the figure below, three circles are tangent to each other and to line L. The radius of circle A is equal to 10 cm and the radius of circle B is equal to 8 cm. Find the radius of circle C.In the figure below, three circles are tangent to each other and to line L. The radius of circle A is equal to 10 cm and the radius of circle B is equal to 8 cm. Find the radius of circle C
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Verified answer
★ 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 :-
______________________________________
★ Solution :-
Given,
» Number of persons to be selected = 3
» Total number of males = 5
» Total number of females = 4
We know that,
And,
So,
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.
Here we have combined combinations for both male and female.
Since 0! = 1
[5! = 5×4×3×2×1 = 120, 3! = 3×2×1 = 6, 2! = 2×1 = 2]
>> 2 male and 1 female ::
Again using same process, we get
[4! = 4×3×2×1 = 24, 1! = 1]
>> 1 male and 2 female ::
Again using the same process, we get
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
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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
>> 1 male and 1 female
>> 0 male and 2 female
Now adding the total number of combinations we get,
→ Total number of combinations = 6 + 16 + 6
→ Total number of combinations = 28
B.) Excluding 0 male cases .
We see that the zero male case is only when the the members are 3 females and 0 males.
So, in first part, we didn't include any 0 male cases.
→ Total number of combinations = 80
In the figure below, three circles are tangent to each other and to line L. The radius of circle A is equal to 10 cm and the radius of circle B is equal to 8 cm. Find the radius of circle C.In the figure below, three circles are tangent to each other and to line L. The radius of circle A is equal to 10 cm and the radius of circle B is equal to 8 cm. Find the radius of circle C
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