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In a group of 6 boys and 4 girls, four children are to be selected.

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In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 30 Jun 2018, 10:54
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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?

A. 159
B. 208
C. 209
D. 212
E. 215

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Re: In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 30 Jun 2018, 11:00
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QZ wrote:
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?

A. 159
B. 208
C. 209
D. 212
E. 215


(At least 1 boy) = (Total) - (0 boys, so all girls)

(At least 1 boy) = 10C4 - 4C4 = 10!/(6!4!) - 1 = 209.

Answer: C.
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Re: In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 30 Jun 2018, 19:38
10 C 4
1 way to choose no boys
10 C 4 - 1
Answer C. 209
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Re: In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 04 Jul 2018, 19:25
QZ wrote:
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?

A. 159
B. 208
C. 209
D. 212
E. 215


We can use the equation:

Total number of ways to select a group of 4 children - number of ways to select a group with no boys = number of was to select a group with at least 1 boy

Total number of ways to select a group of 4 children is 10C4:

10C4 = 10!/(4! x 6!) = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 10 x 3 x 7 = 210

Number of ways to select a group with no boys (i.e., all girls) is 6C0 x 4C4 = 1 x 1 = 1.

Thus, the number of was to select a group with at least 1 boy is 210 - 1 = 209.

Answer: C
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Re: In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 07 Jul 2018, 02:34
Bunuel wrote:
QZ wrote:
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?

A. 159
B. 208
C. 209
D. 212
E. 215


(At least 1 boy) = (Total) - (0 boys, so all girls)

(At least 1 boy) = 10C4 - 4C4 = 10!/(6!4!) - 1 = 209.

Answer: C.


Thank you for the explanation Bunuel! :-) However, could you please explain what is wrong in the below approach:

atleast one guy - 6C1 multiplied by 9C3 - ways to select three people from the remaining 9.

The above approach gives a wrong answer but i am unable to identify the error in logic. Please help!
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Re: In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 07 Jul 2018, 08:40
agarwalaayush2007 wrote:
Bunuel wrote:
QZ wrote:
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?

A. 159
B. 208
C. 209
D. 212
E. 215


(At least 1 boy) = (Total) - (0 boys, so all girls)

(At least 1 boy) = 10C4 - 4C4 = 10!/(6!4!) - 1 = 209.

Answer: C.


Thank you for the explanation Bunuel! :-) However, could you please explain what is wrong in the below approach:

atleast one guy - 6C1 multiplied by 9C3 - ways to select three people from the remaining 9.

The above approach gives a wrong answer but i am unable to identify the error in logic. Please help!


This way will give you duplication.

Say, the boys are {1, 2, 3, 4, 5, 6} and girls are {a, b, c, d}.

6C1 gives you one boy from six, say {1}.
9C3 gives you three children from nine, say {2, 3, a}.
So, the group is {1, 2, 3, a}.

But we will get the same group in other ways too. For example:
6C1 can give you one boy from six, say {2}.
9C3 can give you three children from nine, say {1, 3, a}.
So, the group is {1, 2, 3, a}.

As you can see we got the same group of children.
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In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 15 Jul 2018, 07:06
One can easily grasp these kinds of questions simply by imagining the concept on her/his own. What does it mean to have at least 1 boy? It means that the problem talks about 4 possible scenarios: 1 boy 3 girls, 2 boys 2 girls, 3 boys 1 girl, and 4 boys and no girl.
So,
1st scenario: 6C1 * 4C3
2nd scenario: 6C2 * 4C2
3rd scenario: 6C3 * 4C1
4th scenario: 6C4 * 4C0
And finally, add these 4 and get the answer.
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Re: In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 21 Jul 2018, 06:49
Bunuel wrote:
QZ wrote:
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?

A. 159
B. 208
C. 209
D. 212
E. 215


(At least 1 boy) = (Total) - (0 boys, so all girls)

(At least 1 boy) = 10C4 - 4C4 = 10!/(6!4!) - 1 = 209.

Answer: C.


Why we used 4C4 and not 4!. Isn't 4! ways to choose 4 people from a pool of the 4 people?
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Re: In a group of 6 boys and 4 girls, four children are to be selected.  [#permalink]

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New post 21 Jul 2018, 06:51
AkshdeepS wrote:
Bunuel wrote:
QZ wrote:
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?

A. 159
B. 208
C. 209
D. 212
E. 215


(At least 1 boy) = (Total) - (0 boys, so all girls)

(At least 1 boy) = 10C4 - 4C4 = 10!/(6!4!) - 1 = 209.

Answer: C.


Why we used 4C4 and not 4!. Isn't 4! ways to choose 4 people from a pool of the 4 people?


No. 4! is the number of permutations of 4 distinct items.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In a group of 6 boys and 4 girls, four children are to be selected. &nbs [#permalink] 21 Jul 2018, 06:51
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