There are 13 members in a committee having 7 men and 6 women. In how many ways a team of 4 persons c

Supriya Mondal   9 years AGO

There are 4 ways by which we can form a team
1 man and 3 women=7C1*6C3=140
2 men and 2 women=7C2*6C2=315
3 men and 1 woman=7C3*6C1=210
4 men=7C4=35
So, total=140+315+210+35=700
So, option 7)700 is the right answer

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viswambher   9 years AGO

since there must be atleast 1 men in the team
so the possibilities are

1 men and 3 women
2 men and 2 women
3 men and 1 women
4 men and o women

so this can be done using combinations such as ncr
ncr=(n!/[(n-r)!r!])

the no of ways of selecting 1 man and 3 women or 2men and 2 women or 3men and 1 woman or 4 men and 0 women is

7c1*6c3 + 7c2*6c2 + 7c3*6c1 + 7c4*6c0

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Ramdas Singathiya   9 years AGO

total men=7
total women=6
select 1 man and 3 women= (7c1) *(6c3) = 140
select 2 man and 2 women= (7c2) *(6c2) = 315
select 3 man and 1 women= (7c3) *(6c1) = 210
select 4 man and 0 women= (7c4) *(6c0) = 35

so total ways=700

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ABHINAV PUSHPAM   9 years AGO

=> 7c1*6c3 + 7c2*6c2 + 7c3*6c1 + 7c4
=> 140+315+210+35 = 700
So, answer is 700
i.e. option 7)

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Surit Datta   9 years AGO

7C4 + (7C3*6C1) + (7C2*6C2) + (7C1*6C3) = 700
so option 7
that's it.

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Deepthi   9 years AGO

(7C1*6C3)+(7C2*6C2)+(7C3*6C1)+7C4 =700

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Aashish gohel   9 years AGO

there is at least one men in committee so the no of ways will b
(7C1)(6C3)+(7C2)(6C2)+(7C3)(6C1)+(7C4)(6C0)=700

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aman aditya   9 years AGO

140+315+210+35=700

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W J Krishna Teja   9 years AGO

total no of ways=7c1*6c3+7c2*6c2+7c3*6c1+7c4

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Venkatesh kadali   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4*6c0=700

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Karthick Raja   9 years AGO

TOTAL POSSIBILITIES
13!/(4!*9!)=715 (4 MEMBERS IN A GROUP)
WE CAN CALCULATE how many ways a team of 4 persons can be selected in such a way that at least 1 men would be there in the team SO WE SUBTRACT ALL 4 MEMBERS ARE WOMEN FROM TOTAL POSSIBILITIES


POSSIBILITIES OF ALL 4 ARE WOMEN
6!/(4!*2!)=15

POSSIBILITIES OF a team of 4 persons can be selected in such a way that at least 1 men would be there in the team =715-15=700

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PRASHANT KUMAR   9 years AGO

ANS=7c4*6c0+7c3*6c1+7c2*6c2+7c1*6c3=700

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SHAIK RAFFEE   9 years AGO

the combination of 4 members out o 7 men and 6 women such that atleast one man should present is
7c1*6c3 + 7c2*6c2 +7c3*6c1 +7c4*6c0
=140+315+210+35
=700

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Yathiraj   9 years AGO

13C4 => 715
Total 715 combinations of 4 people committee can be done from group of 13 people.

Combinations of only women
6C4 => 15

So, the committe with at least one men is 715 - 15 = 700.

Option 7.

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MURUGESAN.P   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4*6c0=700

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nikhil kumar   9 years AGO

7C1.6C3 + 7C2.6C2 + 7C3.6C1 + 7C4 = 700

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naina shukla    9 years AGO

we ve to select 4 person out of 13.
there will be atleast 1 men i.e. at most 4 men can be there
There are 7 men and 6 women. so
(7c1*6c3)+(7c2*6c2)+(7c3*6c1)+(7c4*6c0)
=(7*20)+(21*5)+(35*6)+(35)
=700
so option 7 is correct.

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Ayush Gupta   9 years AGO

7C1*6C3+7C2*6C2+7C3*6C1+7C4*6C0=700

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sompon temtheerakij   9 years AGO

1man+3women = 140 ways
2men+2women = 315 ways
3 men+1women = 210 ways
4men+0women = 35 ways
total ways = 140+315+210+35 = 700 ways

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Meenakshi   9 years AGO

=>7C1*6C3 + 7C2*6C2 + 7C3*6C1+7C4*6C0
solving this answer is 700

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Manoj   9 years AGO

4 members in group but at least one man combination is
7c1*6c3 + 7c2*6c2 +7c3*6c1 + 7c4*6c0
=140+315+210+35
=700

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muthu   9 years AGO

7c1*6c3 +7c2*6c2 +7c3*6c3 + 7c4 =700

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B.venkata suresh   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4=700

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Mehul D Parmar   9 years AGO

here in a question ...atleast 1 man would be there in the team then we take 1 man or more than one like 2,3,and 4 also ...than

7 6 7 6 7 6 7
= c * c + c * c + c * c + c
1 3 2 2 3 1 4

=140+210+315+35

=700

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khaleelulla   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4

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Gaurav Malik   9 years AGO

it is pretty simple !
as it is given in the question atleast one man should be there among 4 persons so you can calculate the ans as:
7C1*6C3+7C2*6C2+7C3*6C1+7C4*6C0=700

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Devendra Marghade   9 years AGO

From 7 men and 6 women, atleast 1 men among 4 can be selected as (1M,3W) or (2M,2W) or (3M,1W) or (4M,0W)
So (7C1*6C3) + (7C2*6C2) + (7C3*6C1) + (7C4*6C0)
=140+315+210+35
=700
So answer is option (7)

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akas   9 years AGO

The No Of all the combinations are =n!/(n-r)!r!
therefore solution:
=7!/6! 1! *6!/3! 3! +7!/5! 2! *6!/4! 2! +7!/4! 3!*6!/5! 1! +7!/3! 4!*6!/6! 0!
= 7*20+21*15+35*6+35
=700

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keerthi   9 years AGO

7c1+6c3+7c2*6c2+7c3*6c1+7c4+6c1
700

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venkatesh   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4*6c0

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Krishan Kumar   9 years AGO

atleast 1 men
fr 1 man 3 women=7C1*6C3
fr 2 men 2 women= 7C2*6C2
fr 3 men 1 women= 7C3*6C1
fr 4 men 0 women= 7C4*6C0

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Gowtham   9 years AGO

7c1*6c3 + 7c2*6c2 + 7c3*6c1 + 7c4*6c0 = 700.

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saranya.s   9 years AGO

7 men and 6 women.
A team 4 persona can be selected, such that
NC base R combination
700 ways are there

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NAVEEN   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4=700

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p.supriya   9 years AGO

7c1*6c3+7c2*6c2+7c3*6c1+7c4
140+315+210+35
=700

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Pavithra    9 years AGO

atleast one men must be in the team.So 7c4+(7c3*6c1)+(7c2*6c2)+(7c1*6c3)

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kousik tankala   9 years AGO

total(715)- 0 boys case(15)=700

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Deepak Joshi   9 years AGO

140 + 315 + 210 + 35

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anuradha   9 years AGO

7C1*6C3+7C2*6C2+7C3*6C1+7C4

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Rajat Jain   9 years AGO

(7c1*6c3)+ (7c2*6c2) + (7c3*6c1) + (7c4*6c0)
140+315+210+35
700

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ADERU CHENCHU NITHISH   9 years AGO

no.of ways = 7c4+(7c3*6c1)+(7c2*6c2)+(7c1*6c3)
=35+210+315+140=700

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