[sanjayr84]

case 1: king of spade and a king
(1/52)*(3/51)

Case 2: non-king spade and king
(12/52) * (4/51)

Probability: [(1/52)*(3/51)] + [(12/52)*(4/51)] = 51/(52*51) = 1/52

now the other way:

Case 1: non-king spade and king
(12/52) * (4/51)

Case 2: king of spade and a king
(1/52)*(3/51)

Probability = 1/52

Hence, the total probability = 1/52 + 1/52 = 1/26.

Hope this helps:

[medulla]

Originally Posted by vyomb

hey medulla wats the ans for the ques..

Hi..

Sorry people..
My PC was not working.. So could not visit the forum..
The answer is 1/26...

[crack_this_time]

A customer has to select a fruit basket of 20 fruits. Fruits can be mango, bannana or apple. How many ways can he select fruits?

(there can be zero fruits of one kind, key is M+B+A=20)

know this is kinda a easy, pls explain the method.

ciao

[dragonball]

Originally Posted by sanjayr84

case 1: king of spade and a king
(1/52)*(3/51)

Case 2: non-king spade and king
(12/52) * (4/51)

Probability: [(1/52)*(3/51)] + [(12/52)*(4/51)] = 51/(52*51) = 1/52

now the other way:

Case 1: non-king spade and king
(12/52) * (4/51)

Case 2: king of spade and a king
(1/52)*(3/51)

Probability = 1/52

Hence, the total probability = 1/52 + 1/52 = 1/26.

Hope this helps:

i dont exactly get this solution. Why do we have to consider "now the other way:" ??

[anil_sharma]

Originally Posted by crack_this_time

A customer has to select a fruit basket of 20 fruits. Fruits can be mango, bannana or apple. How many ways can he select fruits?

(there can be zero fruits of one kind, key is M+B+A=20)

know this is kinda a easy, pls explain the method.

ciao

We know that non negative integral solution of the equation:
x1 + x2 + x3+ ....xr = N
are (N+r-1)C r-1

Answer in this case is 22C2 = 231.

[niits]

Hi friends,

Kindly help me solving these questions.

Thanks & regards,
niti


Q2)Seven blank spaces, which are numbered from 1 to 7,are aranged in a row in ascending order from left tp right.Two A's,three B's ,two C's are to b filled inthese seven blank spaces such that the same alphbet never occupy consecutive blank spaces.In how many ways can this be done if the B is always filled in the first blank space?

[crack_this_time]

How many 5 digit numbers can be formed such that it has the following properties:
a)It has at least one zero and atmost three zeros.
b)The non zero digits are not repeated.

we have to find out for 3 cases
Case1: exactly 1 zero
Case2: exactly 2 zero
Case3: exactly 3 zero

what i dont seem to understand is the way the total number of such numbers is calculated.

take Case1:
step1) the first number (ten thousand's position) of the five digits can be placed in 9 ways (any number from 1 to 9)
step2) the second number(thousand's position) can be any number from 0 to 9 (except the one that has been putin the step1.) : total 9 ways
step3) the third number (hundreds position) 8 ways

step4) 7 ways
step 5) 6 ways

but where are we considering the 1 zero thats there in the number.

i m confused :(

[esh.nil]

Originally Posted by niits

Hi friends,

Kindly help me solving these questions.

Thanks & regards,
niti


Q2)Seven blank spaces, which are numbered from 1 to 7,are aranged in a row in ascending order from left tp right.Two A's,three B's ,two C's are to b filled inthese seven blank spaces such that the same alphbet never occupy consecutive blank spaces.In how many ways can this be done if the B is always filled in the first blank space?

@niti...Dude the answer to this qn is 20... the answer has been posted with the approach in the concepts-total fundas thread...

[esh.nil]

Originally Posted by crack_this_time

How many 5 digit numbers can be formed such that it has the following properties:
a)It has at least one zero and atmost three zeros.
b)The non zero digits are not repeated.
i m confused :(

The 3 cases are perfectly right,
1 zero:
we can select the 4 nos in 9*8*7*6 the zero can occupy any 4 places except the first digit, (4c1-- selecting 1 place from 4 available places)
so total no of ways = 9*8*7*6*4 = 12096

2 Zeros:
we can select the 3 nos in 9*8*7 now we have 5 places of which the first place cannot be occupied with zero. so eff 4 places and 2 zeros can be filled in those 2 places in 4c2 way
total no of ways = 9*8*7*6 = 3024

3 Zeros:
we can select the 2 nos in 9*8 ways and the 4 places can be filled with 3 zeros in 4c3 = 4 ways
total nos = 9*8*4 = 288

total nos = 12096+288+3024 = 15048
Is that correct???

[brigadier]

Originally Posted by esh.nil

The 3 cases are perfectly right,
1 zero:
we can select the 4 nos in 9*8*7*6 the zero can occupy any 4 places except the first digit, (4c1-- selecting 1 place from 4 available places)
so total no of ways = 9*8*7*6*4 = 12096

2 Zeros:
we can select the 3 nos in 9*8*7 now we have 5 places of which the first place cannot be occupied with zero. so eff 4 places and 2 zeros can be filled in those 2 places in 4c2 way
total no of ways = 9*8*7*6 = 3024

3 Zeros:
we can select the 2 nos in 9*8 ways and the 4 places can be filled with 3 zeros in 4c3 = 4 ways
total nos = 9*8*4 = 288

total nos = 12096+288+3024 = 15048
Is that correct???

you forgot to multiply by 4!, 3!, 2! in cases 1,2,3 resp