Question

There are 10 questions, each question is either True or False. Number of different sequences of incorrect answers is also equal to
A) The number of ways in which a normal coin tossed 10 times would fall in a definite order if both the Heads and Tails are present.
B) The number of ways in which a multiple choice question containing 10 alternatives with one or more than the correct alternatives can be answered.
C) The number of ways in which it is possible to draw a sum of money with the 10 coins of different denominations taken some or all at a time.
D) The number of different selections of 10 indistinguishable things taken some or all at a time.

Answer 1

In this question, we have to find the number of ways for each given option and the given statement. We must know the number of items and the number of ways in which each item can be selected. Then, multiply the number of ways in which each item is selected by the total number of items in order to find the total number of possible ways. Also, subtract the number which contradicts the given condition from total ways possible to get the required answer.

Complete step by step solution:
We are given that there are 10 questions that need to be answered, either true or false.
So, there are 2 ways in which each question can be answered.
Hence, the number in which 10 questions can be answered will be given by:
$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}$
But, one among the possible sequences will be correct. So, we will subtract 1 from all the ways that are possible.
Hence, the number of different sequences of incorrect answers is also equal to ${2^{10}} - 1$.

Next, we calculate the number of ways in which a normal coin tossed 10 times would fall in a definite order if both the Heads and Tails are present.
Since a coin is tossed 10 times and there are only 2 possibilities, that is, head or tail.
Therefore, the number of ways in which a normal coin tossed 10 times will be$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}$.
But, if we want cases only when both Heads and Tails are present.
There is one case when only heads will be present and one case when only tails will be present.
So, we will subtract 2 from total ways possible.
Hence, the number of ways in which a normal coin tossed 10 times would fall in a definite order if both the Heads and Tails are present are ${2^{10}} - 2$
For option B, calculate the number of ways in which a multiple choice question containing 10 alternatives with one or more than the correct alternatives can be answered.
As there is 1 question and there are 10 options.
Either the option is selected or not selected. Hence, there are 2 possible cases for each option.
Hence, the number of all possible ways are $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}$
But, there will be one when no option is marked and we have to answer by marking at least one option.
Therefore, the number of ways in which a multiple choice question containing 10 alternatives with one or more than the correct alternatives can be answered are ${2^{10}} - 1$.

In option C, we have 10 coins and we have to choose some or all of them.
Now, either each coin can be selected or not selected.
Hence, the number of all possible ways are $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^{10}}$
But, there will be one case when no coin is selected. So, subtract 1 from all possible cases
Therefore, the number of ways in which it is possible to draw a sum of money with the 10 coins of different denominations taken some or all at a time are ${2^{10}} - 1$.

In option D, there are 10 same items. Hence, there has to be no choice.
We can take one item, two items and so on till 10 items.
Hence, the number of different selections of 10 indistinguishable things taken some or all at a time is 10.
Therefore, option B and C are correct.

Note:
While calculating the solution, one must not forget to subtract the number of cases which violates the given condition. For better understanding, there are two cases when only one of the heads or tails are present which are when only Heads are present or only Tails are present in option B. Also when the things are the same, there are no choices as in option D.