Question

In how many ways can Ravi divide 11 identical chocolates between 3 of his friends
(excluding Ravi) such that each friend gets an odd number of chocolates and every
friend gets at least one chocolate.
​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Answer:

The correct answer is 15.

Step-by-step explanation:

i) There are four possible combinations in which 11 chocolates can be divided in three friends such that each one gets odd number of chocolates and everyone gets at least one chocolate:

(1,1,9) ; (1,3,7) ; (1,5,5) ; (3,3,5)

ii) To calculate the possible number of permutations in which each combination can be distributed, we can use following formula:

Permutations of n objects with no repetition: n!

Permutations of n objects with r repetitions: n!/r!

iii) Hence, for combinations with 2 repeated numbers i.e. (1,1,9), (1,5,5),(3,3,5), the total number of possible permutations are:

3 x (3!/2!) = 3 x 3 = 9

For combination with no repeated number i.e. (1,3,7), the total number of possible permutations are:

3! = 6

iv) Total number of possible permutations= 9+6= 15

v) Hence, the total number of ways in which Ravi can divide 11 identical chocolates between 3 friends such that everyone gets odd number of chocolates and every friends gets at least one chocolate is 15.

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