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 numberof ways |
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) | |||||||||||||||||||
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) | ||||||||||||||||||||
|
||||||||||||||||||||
| = (24 + 90 + 80 + 15) | ||||||||||||||||||||
| = 209. |
Discussion:
92 comments Page 2 of 10.
Mohit said:
1 decade ago
Can anyone please help me to reduce the time spent in lengthy calculations. I understood how it can be solved. But calculations take so much time and we don't have much time in solving one particular question.
If I try to solve fast, small silly mistake happens and could not get answer in 1st attempt which increases the time spent to re-check it.
If I try to solve fast, small silly mistake happens and could not get answer in 1st attempt which increases the time spent to re-check it.
Anusha said:
1 decade 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).
Why to change only the first term and last term with formula. Why don't the formula 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).
Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain?
Prabhu said:
1 decade 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).
Why to change only the first term and last term with formula. Why don't the formula 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).
Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain?
Kiran said:
1 decade ago
There is a trick here just oppose the question like following:
We have 6 Boys and 4 Girls,
1) Total No. of Possible ways = 10c4 -->210 (selecting 4 out of 10 Children).
2) Tricky here(Oppose) select no boy at all i.e select 4 children from only girls = 4C4 --> 1.
Finally : Subtract 2nd from 1st 210-1 = 209 Answer.
We have 6 Boys and 4 Girls,
1) Total No. of Possible ways = 10c4 -->210 (selecting 4 out of 10 Children).
2) Tricky here(Oppose) select no boy at all i.e select 4 children from only girls = 4C4 --> 1.
Finally : Subtract 2nd from 1st 210-1 = 209 Answer.
Rishi Jain said:
10 years ago
Hi @Aditya,
In 1st step : ......+ (6c1 x 4c3) +......
In 2nd step : ......+ (6c1 x 4c1) +...... How ?
You say that,
Its because nCr = nC (n-r).
So 4C3 = 4C (4-3) = 4C1.
Its one of the formula. But I have confused.
That what about 6c1 we should also do 6c1 = 6c (6-1) = 6c5.
But we didn't know why? Please explain me.
In 1st step : ......+ (6c1 x 4c3) +......
In 2nd step : ......+ (6c1 x 4c1) +...... How ?
You say that,
Its because nCr = nC (n-r).
So 4C3 = 4C (4-3) = 4C1.
Its one of the formula. But I have confused.
That what about 6c1 we should also do 6c1 = 6c (6-1) = 6c5.
But we didn't know why? Please explain me.
Heenu said:
7 years ago
Easiest way:
Since 4 students are to be selected and there are 4 girls, hence we don't want a combination consisting of all girls: 4c4 =1.
Total no of combinations available: (Total students= 6 boys + 4 girls =10 and 4 need to be selected) 10c4= 210.
Hence favorable combinations = 210 -1 = 209.
Since 4 students are to be selected and there are 4 girls, hence we don't want a combination consisting of all girls: 4c4 =1.
Total no of combinations available: (Total students= 6 boys + 4 girls =10 and 4 need to be selected) 10c4= 210.
Hence favorable combinations = 210 -1 = 209.
Phanichander said:
1 decade ago
Hi it is simple, read question ones again
We have 6-boys and 4-girls we have to select only 4 members from the groups so in that once condition least 1 boy means one boy or more 1 than boy or all the boys only
(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)=209
We have 6-boys and 4-girls we have to select only 4 members from the groups so in that once condition least 1 boy means one boy or more 1 than boy or all the boys only
(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)=209
Himanshu Kulkarni said:
9 years ago
The question says at least one boy is to be selected.
Selecting 1 boy from 6 boys is 6C1 i.e. 6 ways.
And from remaining 9 children 3 are to be selected randomly (any combination of girls and boys) which is 9C3.
So why not 6C1 x 9C3 matches the answer?
Selecting 1 boy from 6 boys is 6C1 i.e. 6 ways.
And from remaining 9 children 3 are to be selected randomly (any combination of girls and boys) which is 9C3.
So why not 6C1 x 9C3 matches the answer?
Jigs said:
1 decade ago
Another Shortcut Method:.
Total student=6+4=10 The number of ways to selecting 4 girls=10c4, The number of ways to selecting no boys= 4c4 because total boys are=6 so 10-6=4 to select 4 children so, 4c4 so selecting at-least one boy 10c4-4c4=210-1=209.
Total student=6+4=10 The number of ways to selecting 4 girls=10c4, The number of ways to selecting no boys= 4c4 because total boys are=6 so 10-6=4 to select 4 children so, 4c4 so selecting at-least one boy 10c4-4c4=210-1=209.
Sandeep said:
3 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?
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