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Aptitude - Permutation and Combination - Discussion
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Discussion Forum : Permutation and Combination - General Questions (Q.No. 6) Permutation and Combination 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? 159 194 205 209 None of these
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 | = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) |
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) |
| = (6 x 4) + |  | 6 x 5 | x | 4 x 3 |  | + | | 6 x 5 x 4 | x 4 | | + | | 6 x 5 | | | 2 x 1 | 2 x 1 | 3 x 2 x 1 | 2 x 1 | |
| = (24 + 90 + 80 + 15) |
| = 209. |
Discussion:
92 comments Page 1 of 10. Newest
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Gayathri m said:
5 months ago The second step is not determined.i.e (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Divyesh Khunt said:
1 year 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.
(25) Prajwal BK said:
2 years ago 1 case is missed out i.e., 4 girls can also be one more case.So that makes up to 5 possible scenarios.
(6) Ritesh Sadh said:
2 years 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.
(10) ATM said:
3 years 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 boyHence 210-1 = 209.
(31) Nilesh said:
3 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) +...6C4In 2nd step : ... + (6c1 x 4c1) +...6C2 How ?Why don't the formula nCr = nC(n-r). So 4C3 = 4C(4-3) = 4C1 apply to the other terms. Please Explain.
(3) Sandeep said:
4 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?
(2) Sayuj said:
4 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.
(1) Ezhil said:
5 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 (n-r), 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 (n-r) 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:
5 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.
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