IIT JEE , AIEEE Forum - Mathematics , Algebra

avik987

a family consists of a grandfather , 5 sons and 8 grandchildren. they r to be seated in a row for a dinner. the grandchildren wish to occupy the four seats at each end of the row and the grandfather refuses to have a grandchild besides him. In how many ways can the family be made to sit ?

my solution.............

total no of seats = 14

8 grandchildren can be made to seat in 8! ways .

grand father can be made to sit in 5 - 1 ways.......

and the remaing seat can be filled in 5! ways.

so total = 8! * 4 * 5!

but ans is 4! * 4 * 5! plzzz help out....................

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in how many ways can a mixed doubels game of tennis can be arranged from 5 married couples, if no husband and wife play in the same game?

my solution..........

let two teams be A n B

for team A 5 husbands can be selected in 5 ways and 4 wifes can be selected in 4 ways........

for team B 4 husbands can be selected in 4 ways n 3 wifes can be selected in 3ways

so total ways is 5*4*4*3 = 240

but answer in 60

plzzzzz guys help out...........

uday_zingtudor

Hey in ur second question,

No couple must play in the same game means the wife and hubby should be neither on the same nor on the opposite side.

Select two wives from the total 5 in 5C2 ways.Now u can't select their hubbies.So u can select only 3 of them in 3C2 ways.Now u can interchange the couples selected for match.So multiply by 2.

So 5C2 x 3C2 x 2 = 60

uday_zingtudor

edit:
I have doubt for the first one man as the children can seat themselves in 8! ways. Except that everything is same as the solution below.

akshay.khare91

solutions...

1.

we have 14 seats -- | | | | | | | | | | | | | |

in the four seats form last and starting the grandchildren will sit
and they can arrange themselves in 4! ways...

now grandfather donot want to sit beside the grandchildren hence
he cannot sit on 5th and 10th chair (COUNT CHAIR FROM STARTING)

hence he can select 1 chair from 4 chairs (i.e. 6TH 7Th 8Th 9TH)
this can be done in 4C1 ways = 4

and now on the remaining 5 chairs the sons can arrange thenselves
in 5 ! ways

so ans . = 4! * 4 *5! ...

2 ..

first we select 2 men from 5 this can be done in 5C2 ways

now if we want to select wifes we have only three options
because we cannot select the wifes of 2 man to whom we have
selected ...
hence we have 3 options = 3C2

now we can arrange them in 2 ways
M1W1 M2W2 and M1W2 M2W1

hence ans. is 5C2 * 3C2 * 2 = 60

hope u got it !!!!
if not nudge me!!!!!

avik987

plz xplain wy it shud be 4! instead of 8!

dats wherre my doubt is...........

avik987

wats wrong with my solutions for d second problem ????

can neone xplain????????

vineetnegi

your second part can be done easily

x.............female

y.................male

your ans ;

coefficient of x^2y^2

(1+x+y)^5

by multinomial theorem

5!/2!*2!

=30

now 2 ways are ppossible of making teams of 2 male and 2 female

so

30*2=60

i posted this earlier but in your other post

vineetnegi

your method is not wrong but just a little

mistake

you siad that no couple come in one match

you proceeded correctly

5............men

5..............women

so no of ways

5*4

correct

for selecting second team

you chose

4 men but you forgot that the wife you selected in first team has his husband out in the group still so you can ttake that men

4-1

no you canchoose 3 men

same for the men you choose first team

his wife is out

so 4-1=3

now no of way is 3-1=2 ...........leaving the wife of hus. selected insecond team

5*4*3*2=120

but you see that teams can be interchanged

120/2=60

cheer up!!!!

avik987

but wat about my first question?????????//

no xplanation for it yet?????????

RyuAmakusa

avik987 i think your ans. is correct for the first q.

ayshwarya

fr d 1st que
d total no. of seats reqd at d table is 14
d grand children can occupy d 4 seats on either side of d table

in 8P8 ways d grand father can occupy a seat in 4P1 ways i.e
(5-1)P1 ways
d remaining seats can b occupied in 5P5 ways der4
d reqd no. ways is
8 ! x 5 ! x4 ways

avik987

for d first one it seems like d ans shud be 8! * 4 * 5!

but d ans is 4! * 4 * 5!

plzzz zplain guys...........

no satisfactory reply yet........

its an arihant algebra xample.........p n c

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