IIT JEE , AIEEE Forum - Mathematics , Algebra - TOUGH OBJ QUESTIONS-P&C--fast replies plz

warrior

1)6 DIFFERENT GIFTS ARE TO BE GIVEN TO 9 STUDENTS SO THAT 4 SPECIFIED STUDENTS SHUD GET ATLEAST 1 GIFT EACH=

ANS:17760

2) a,b,c are integers greater than 1 .if abc= 2^4.3^5.5^3 (^ means power) then possible no.of ordered triple (a,b,c)=

ans:2793

warrior

explanation plz...anyone

akhil_o

1)the 4 specified students get 1 prize each initially...
so we have
6C4 ways of distributing them...
now we have 2 gifts remaining...each can go to any of 9 students
so 9*9 ways

so total
=6C4*9*9
=1215

bvsatyaram

Algorithm for the first question:

The students for whom there must be atleast 1 gift given are specified. So u dont need to select these 4 students. So, 6C4 doesnt come into picture.

The algorith to solve the problem is as follows:
- First distribute 9 gifts among 6 students without any restrictions
- then subtract the cases where atleast one of the 4 specified students dont get a gift. For this case, u need to use principle of inclusion and exclusion

Cheers

bvsatyaram

Algorithm for second problem:

the problem can be made simple by interpreting it in a different perspective.
Consider a, b, c as there different boxes. U have 4 red ball (red is the divisor 2 in the product), 5 green balls (green is 3) and 3 blue balls(blue is 5).

Now the problem is to place these 12 balls in 3 boxes, so that each box is filled with atleast one ball. Once again principle of inclusion and exclusion... So, the problem is same as the previous one..

pollp2

yes i got the ans but i'll post the explanation tommorow

vineetnegi

to your second question the ans is

you can divide the given expression into three no. product by using general sub division rule

(4+2)C2 *(5+2)C2 *(3+2)C2

=3150

but this ans contains both 1 and greater than 1 triplets

suppose that one variable is a=1

then in how many ways can you distribute given expression into two factor product(4+1)C1 *(5+1)C1*(3+1)C1=120

but this will contain 1 as a factor so sub. 1 for the one permutation in which b=1

119

do it for three var.

119*3=357

ans

3150-357=2793

thevyzz

1)

Arranging the four children assured of a gift first, you get

the following scenarios

1 1 1 1 ( 2 gifts for one child in the five others)

1 1 1 1 ( a gift apiece for two children in the other five)

1 1 1 2 ( 1 gift for others)

1 1 2 2 ( no gift for others)

1 1 1 3 ( no gift for others)

for case 1, no of ways = 5 * 6!/2! = 1800 ( selection of a child from other 5 * no of ways of distributing the different gifts among them)

for case 2, no of ways = 5C2 * 6! = 7200 ( selection of two from the other five * no of ways of distributing the different gifts among them)

for case 3, no of ways = 5 * 4 * 6!/2! = 7200 ( selection of one child from the other five * selection of the child getting two gifts among the assured four * no of ways of distributing the different gifts among them)

for case 4, no of ways = 4C2 * 6!/(2! * 2!) = 1080 ( selection of two children from the assured four to get two gifts each * no of ways of distributing the different gifts among them)

for case 5, no of ways =4 * 6!/3! = 480 ( selection of one child from the assured four to get 3 gifts * no of ways of distributing the different gifts among them)

total no of ways = 17760

vineetnegi

thats exacltly what i thought abt too.............

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