Set 9

From i and ii, the server’s daughter has to be the receiver, and the daughter of both the server and the receiver’s father.

->Server’s daughter/receiver is Charu’s daughter

->Server is Charu

->The person directly across Charu is Charu’s husband

->The person directly across Charu’s daughter is Charu’s father.

 

The arrangement can be represented as under:

Charu (Server)	Charu’s Father
Charu’s Husband	Charu’s Daughter (Receiver)

 

Thus, the statements are made by Charu’s father

Let M indicate a man and W indicate a woman. Following are the possible arrangements:

I.

MMMWWW

1  2 3 4 5 6

Three men can be interviewed in the first 3 slots in 3*2*1 = 6 ways; three women can be interviewed in the next three slots in 3*2*1 = 6 ways.

->Total number of ways in which the candidates can be interviewed = 6*6 = 36

 

II.

MWMWMW

1 2  3 4 5 6

Interviews at slots 1 & 2 can be conducted in 3 ways (M₁W₁, M₂W₂, M₃,W₃), at slots 3 & 4 in 2 ways (two out of the remaining), and at slots 5 & 6 in 1 way.

->Total number of ways in which the candidates can be interviewed = 3*2*1 = 6

 

III.

MMWMWW

1  2 3 4 5 6

Interview at slot 1 can be conducted in 3 ways (M₁, M₂, M₃), at slot 2 in 2 ways (two of the remaining men), at slot 3 in 2 ways (corresponding to the two men in slots 1 & 2), at slot 4 in 1 way (the person remaining out of M₁, M₂, M₃), at slot 5 in 2 ways (corresponding to any two of the remaining women), and at slot 6 in 1 way (the women who hasn’t been interviewed till now)

->Total number of ways in which the candidates can be interviewed = 3*2*2*1*2*1 = 24

 

IV.

MWMMWW

1  2 3 4 5 6

Interview at slot 1 can be conducted in 3 ways (any one of the three men), at slot 2 in 1 way (corresponding to the man who is interviewed at slot 1), at slot 3 in 2 ways (two of the remaining men), at slot 4 in 1 way (the man who yet remains), at slot 5 in 2 ways (any of the two remaining women), and at slot 6 in 1 way (the remaining woman)

->Total number of ways in which the candidates can be interviewed = 3*1*2*1*2*1 = 12

 

V.

MMWWMW

1  2 3 4 5 6

Interview at slot 1 can be conducted in 3 ways (any one of the three men), at slot 2 in 2 ways (two of the remaining men), at slot 3 in 2 ways (corresponding to the two men in slots 1 & 2), at slot 4 in 1 way (corresponding to the man whose counterpart is not interviewed at slot 3), at slot 5 in 1 way (the man who yet remains), and at slot 6 in 1 way (corresponding to the man who is interviewed

at slot 5)

->Total number of ways in which the candidates can be interviewed = 3*2*2*1*1*1 = 12

->Total number of ways in which the candidates can be interviewed = I+II+III+IV+V = 36+6+24+12+12 = 90

 

Note:

• While calculating the number of ways in which the interviews can be conducted for each of the 5 possibilities, we have applied the fundamental principle of counting, which states “If you have an event ‘a’ and another event ‘b’, then the total number of different outcomes for the events is a*b”.
• Finally, we add the number of ways corresponding to each of the 5 possibilities, because all these events are mutually exclusive i.e., one event excludes the possibility of occurrence of others.

The first step here is to eliminate values which are not possible in the grid.

• 0 is eliminated because it will make one of the three triplets zero (n₁ n₂n₃; n₂n₄n₅; n₅n₆n₇), and the other two triplets can’t be zero because we have only one zero in the set of possible values.
• 5 and 7 are eliminated because their occurrence in any of the triplets will make that triplet a multiple of 5 or 7 respectively and the remaining 2 triplets can’t be such multiples because there are no more 5s or 7s available (both 5 and 7 are prime and don’t have factors other than 1)

The remaining digits are: 1, 2, 3, 4, 6, 8, 9 which can be resolved into prime factors in the following ways:

1 -> 1

2 -> 2

3 -> 3

4 -> 2*2

6 -> 2*3

8 -> 2*2*2

9 -> 3*3

Thus, there are seven 2s, four 3s and one 1 in the grid. Out of the three triplets, n₁n₂n₃ and n₅n₆n₇ should have equal number of 2s and 3s if the product is the same, which is possible if each of the triplets has three 2s and two 3s as shown below:

n₁*n₂*n₃ = 2*2*2*3*3

n₅*n₆*n₇ = 2*2*2*3*3

Looking at the available digits we get the following values:

n₁*n₂*n₃ = 4*2*9 —– ❶

n₅*n₆*n₇ = 8*9*1 —– ❷

Now n₄ will have a unique value of 2 because out of seven 2s and four 3s, all are taken but for one 2, which corresponds to the value of n₄. Also, since the triplets have the same value, n₂*n₅ should be equal to 36, as n₂*n₄*n₅ should be equal to 72. From ❶ & ❷, 36 is possible only as 4*9 or 9*4. Thus,

n₂*n₄*n₅ = 4*2*9 OR 9*2*4

Hence, n₂ can have two values i.e. 4 and 9.

Answer = 4, 9

Note:

With the above conditions, eight arrangements are possible as shown-

Possible values for n₁ = 3, 6, 1, 8

Possible values for n₂ = 4, 9

Possible values for n₃ = 6, 3, 1, 8

Possible values for n₄ = 2

Possible values for n₅ = 9, 4

Possible values for n₆ = 8, 1, 3, 6

Possible values for n₇ = 1, 8, 6, 3