Aptitude  Permutation and Combination  Discussion
Discussion Forum : Permutation and Combination  General Questions (Q.No. 6)
6.
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
Answer: Option
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways 
= (^{6}C_{1} x ^{4}C_{3}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{4})  
= (^{6}C_{1} x ^{4}C_{1}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{2})  


= (24 + 90 + 80 + 15)  
= 209. 
Discussion:
91 comments Page 1 of 10.
Divyesh Khunt said:
2 months ago
Another easy method here;
It's given at least 1 boy meaning (>=1) so we can subtract calculate 0 boys selected and subtract it from the total combinations possible.
Combination(boys>=1) =total combination  combination (0boys selected),
= 10C4  6C0 * 4C4.
= 210 1.
= 209.
It's given at least 1 boy meaning (>=1) so we can subtract calculate 0 boys selected and subtract it from the total combinations possible.
Combination(boys>=1) =total combination  combination (0boys selected),
= 10C4  6C0 * 4C4.
= 210 1.
= 209.
(3)
Prajwal BK said:
10 months ago
1 case is missed out i.e., 4 girls can also be one more case.
So that makes up to 5 possible scenarios.
So that makes up to 5 possible scenarios.
(3)
Ritesh Sadh said:
1 year ago
Let's break down the calculation step by step:
Given that there are 6 boys and 4 girls, we need to select 4 children such that at least one boy should be there.
We have four possible scenarios:
1. Selecting 1 boy and 3 girls.
2. Selecting 2 boys and 2 girls.
3. Selecting 3 boys and 1 girl.
4. Selecting 4 boys.
Now, let's calculate the number of ways for each scenario:
1. Selecting 1 boy and 3 girls:
 Number of ways to select 1 boy from 6 boys: 6C1 = 6 (combinations).
 Number of ways to select 3 girls from 4 girls: 4C3 = 4 (combinations).
 Total ways for this scenario: 6C1 x 4C3 = 6 x 4 = 24.
2. Selecting 2 boys and 2 girls:
 Number of ways to select 2 boys from 6 boys: 6C2 = 15 (combinations),
 Number of ways to select 2 girls from 4 girls: 4C2 = 6 (combinations),
 Total ways for this scenario: 6C2 x 4C2 = 15 x 6 = 90.
3. Selecting 3 boys and 1 girl:
 Number of ways to select 3 boys from 6 boys: 6C3 = 20 (combinations),
 Number of ways to select 1 girl from 4 girls: 4C1 = 4 (combinations),
 Total ways for this scenario: 6C3 x 4C1 = 20 x 4 = 80.
4. Selecting 4 boys:
 Number of ways to select 4 boys from 6 boys: 6C4 = 15 (combinations).
Now, add up the total ways for all four scenarios:
Total ways = (1) + (2) + (3) + (4) = 24 + 90 + 80 + 15 = 209.
So, there are 209 different ways to select four children from the group such that at least one boy is included.
Given that there are 6 boys and 4 girls, we need to select 4 children such that at least one boy should be there.
We have four possible scenarios:
1. Selecting 1 boy and 3 girls.
2. Selecting 2 boys and 2 girls.
3. Selecting 3 boys and 1 girl.
4. Selecting 4 boys.
Now, let's calculate the number of ways for each scenario:
1. Selecting 1 boy and 3 girls:
 Number of ways to select 1 boy from 6 boys: 6C1 = 6 (combinations).
 Number of ways to select 3 girls from 4 girls: 4C3 = 4 (combinations).
 Total ways for this scenario: 6C1 x 4C3 = 6 x 4 = 24.
2. Selecting 2 boys and 2 girls:
 Number of ways to select 2 boys from 6 boys: 6C2 = 15 (combinations),
 Number of ways to select 2 girls from 4 girls: 4C2 = 6 (combinations),
 Total ways for this scenario: 6C2 x 4C2 = 15 x 6 = 90.
3. Selecting 3 boys and 1 girl:
 Number of ways to select 3 boys from 6 boys: 6C3 = 20 (combinations),
 Number of ways to select 1 girl from 4 girls: 4C1 = 4 (combinations),
 Total ways for this scenario: 6C3 x 4C1 = 20 x 4 = 80.
4. Selecting 4 boys:
 Number of ways to select 4 boys from 6 boys: 6C4 = 15 (combinations).
Now, add up the total ways for all four scenarios:
Total ways = (1) + (2) + (3) + (4) = 24 + 90 + 80 + 15 = 209.
So, there are 209 different ways to select four children from the group such that at least one boy is included.
(7)
ATM said:
1 year ago
A Group of 4 children, out of 10 children will be;
10c4 = 210.
As 4 girls are there;
So only 1 group is possible without the boy
Hence 2101 = 209.
10c4 = 210.
As 4 girls are there;
So only 1 group is possible without the boy
Hence 2101 = 209.
(25)
Nilesh said:
2 years ago
We may have(1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
In 1st step : ... + (6c1 x 4c3) +...6C4
In 2nd step : ... + (6c1 x 4c1) +...6C2 How ?
Why don't the formula nCr = nC(nr).
So 4C3 = 4C(43) = 4C1 apply to the other terms. Please Explain.
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
In 1st step : ... + (6c1 x 4c3) +...6C4
In 2nd step : ... + (6c1 x 4c1) +...6C2 How ?
Why don't the formula nCr = nC(nr).
So 4C3 = 4C(43) = 4C1 apply to the other terms. Please Explain.
(3)
Sandeep said:
2 years ago
@All.
6C1 x 9C3.
Why can't be true?
For four Places we are taking conformally 1 from boys i.e 6C1 and there are total of 9(boys+girls) from we need to pick 3. So 9C3.
Hence the answer can be: 6C1 x 9C3.
Can anyone explain what's wrong with this?
6C1 x 9C3.
Why can't be true?
For four Places we are taking conformally 1 from boys i.e 6C1 and there are total of 9(boys+girls) from we need to pick 3. So 9C3.
Hence the answer can be: 6C1 x 9C3.
Can anyone explain what's wrong with this?
(2)
Sayuj said:
3 years ago
Why can't we just simply do 6c1*9c3?!
In this, we've included that "minimum" condition of including that 1 boy and the rest 9 kids can be selected at random. why not? Someone, please explain.
In this, we've included that "minimum" condition of including that 1 boy and the rest 9 kids can be selected at random. why not? Someone, please explain.
(1)
Ezhil said:
4 years ago
@Huma. @Imnikesh.
4c3 becomes 4C1, Let me explain 4C3 means we have to take 3 girls out of 4 girls, generally by using the aforementioned formula was nCr = nC (nr), this formula is only should be used on the condition that is value is half of the number greater than n value
i.e. nCr = 4C3, r=3 value which means are value is half of the value of n=4 increased (half of the value of n=4 is 2 I. E r=3 > 2) so that we applying nCr = nC (nr) formula, 4C3 becomes 4C1. On the other case 6C1 we should not be applied this formula because r=1 value is smaller the half of the value of n=6 (half of the value of n=6 is 3 i.e r=1 < 3) that's why we could not be used aforesaid nCr formula.
4c3 becomes 4C1, Let me explain 4C3 means we have to take 3 girls out of 4 girls, generally by using the aforementioned formula was nCr = nC (nr), this formula is only should be used on the condition that is value is half of the number greater than n value
i.e. nCr = 4C3, r=3 value which means are value is half of the value of n=4 increased (half of the value of n=4 is 2 I. E r=3 > 2) so that we applying nCr = nC (nr) formula, 4C3 becomes 4C1. On the other case 6C1 we should not be applied this formula because r=1 value is smaller the half of the value of n=6 (half of the value of n=6 is 3 i.e r=1 < 3) that's why we could not be used aforesaid nCr formula.
(1)
Huma_ said:
4 years ago
Can anyone explain this 2nd step:.
(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Why the nCr formula is applied to only (6C1 x 4C3) & (6C4) and not to all? Please explain.
(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Why the nCr formula is applied to only (6C1 x 4C3) & (6C4) and not to all? Please explain.
(2)
Nithin said:
4 years ago
Why can't we do 6C1*9C3?
First choose a boy from group of 6 and then choose any three from remaining 9 people.
First choose a boy from group of 6 and then choose any three from remaining 9 people.
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