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Question

The number of ways in which 5 different toys can be distributed to 3 children if each child can get any number of toys, is also equal to

A
Number of subsets of A={1,2,3,4,5,6,7,8,9,10} which contain 2 elements of A such that their sum not equal to 11
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B
Number of non-negative integral solutions of the equation xyz=2310
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C
Number of 6digit numbers less than 200000 formed by using only the digits 1,2 and 3
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D
Number of all possible selections of one or more questions from 5 given questions, with an alternative question corresponding to each question
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Solution

The correct options are
A Number of subsets of A={1,2,3,4,5,6,7,8,9,10} which contain 2 elements of A such that their sum not equal to 11
B Number of non-negative integral solutions of the equation xyz=2310
C Number of 6digit numbers less than 200000 formed by using only the digits 1,2 and 3
Number of ways in which 5 different toys can be distributed to 3 children if each child can get any number of toys =35

There are 5 pairs of numbers having sum equal to 11 i.e.,
(1,10);(2,9);(3,8);(4,7);(5,6)
For no two elements of the subset to have sum equal to 11:
For each pair, we can select one number or not select both the numbers. (3 ways)
So, required number of subsets =35

xyz=2310=235711
2 can be distributed to x,y,z in 3 ways.
3 can be distributed to x,y,z in 3 ways.
5 can be distributed to x,y,z in 3 ways
7 can be distributed to x,y,z in 3 ways.
11 can be distributed to x,y,z in 3 ways.
Hence, total number of ways is 35

1,,,,,
1st place can be filled only in one way, i.e., 1
Remaining 5 places can be filled in 35 ways.
Hence, total number of ways =1×35=35

For every question, we have three choices : to do that question or its alternative or not to do the question.
So, number of all possible ways is (3)5 as there are 5 questions and 3 choices for each question.
But we have to select at least one question.
So there will be 351 ways.

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