Time for some examples!
Q1. In how many ways can a selection be made from a lot of 10 distinct objects, such that
(i) any number of objects may be selected?
(ii) atleast 1 object must be selected?
(iii) atmost 3 objects can be selected?
Sol. (i) This one is a direct application of the formula. The required answer is 210 or 1024.
(ii) A small modification in the previous answer. Out of the total number of selections, there will be 1 selection which will be empty. Or, there is only 1 way to select no object. This must be removed from the previous answer. Therefore, the required number of selections will be 210 – 1 = 1023.
(iii) In this case we’ll use the other formula for all possible selections. If we are to select atmost 3 objects, then either we can select no object, or select 1, or select 2, or select 3. The required number of ways will therefore be 10C0 + 10C1 + 10C2 + 10C3.
Let’s also consider the case where the 10 objects are identical
(i) Using the formula for identical objects, the answer will be 10 + 1 or 11.
(ii) Similar to the (ii) above, we have to remove 1 way from the previous answer – there will be 10 possible selections.
(iii) The number of ways to select 0 or 1 or 2 or 3 objects will be 1 in each case. Therefore, the number of ways will be 1 + 1 + 1 + 1 or 4.
Q2. In how many ways can a test paper of 5 questions be attempted, each question being of a TRUE/FALSE type?
Sol. This problem again involves the concept of all possible selections, but in this case after selecting a question we have further choices to make – either mark TRUE or FALSE.
The thinking – ‘either select no question, or select 1 question and mark it TRUE or FALSE, or select 2 questions and mark both of them (in 4 ways)… and so on’
I’ll write the answer straight away – 5C0 + 5C1.2 + 5C2.22 + 5C3.23 + 5C4.24 + 5C5.25, and you do the thinking!
But there is one more way to think about the problem – the one which involved the ‘basket‘.
To attempt the question paper, we’ll begin at the first question. We have three options – mark it TRUE, mark it FALSE, or leave that question. Then we’ll move on the next question, and we have three choices again, and so on till we reach the last question.
Since at every step, we have 3 choices, the number of ways will be 3 x 3 x 3 x 3 x 3 or 35. The complicated expression obtained previously evaluates to the same value which we obtained just now.
Once again, black magic.
Q3. In how many ways can a selection of balls be made from a bag of 5 identical white balls and 4 different colored balls?
Sol. Another direct application of the formula which we had derived in the previous lesson. The required number is (5 + 1)24, or 96.
Q4. In Q3, what will be the number of ways, if
(i) atleast one white ball must be selected?
(ii) atleast one colored ball must be selected?
Sol. (i) Thinking along similar lines, the number of ways to select atleast one white ball will be 5. And since there isn’t any restriction on the number of colored balls, the number of ways to select them will still be 24. Therefore, the total number of ways will be 5 x 24 or 80
(ii) This one is quite similar to the previous one, except that the restriction is on the colored balls. The number of ways to select the white balls will be 6 and that of the colored balls will be 24 – 1 or 15. Therefore, the total number of ways will be 6 x 15 or 90.
Q5. In how many ways can a test paper of 10 questions, divided in to two sections of 4 and 6 questions, be attempted, if one must answer atleast one question from each section?
Sol. This one is again similar to the previous one, but both the lots now consist of different objects. The answer will be (24 – 1)(26 – 1). I hope you got that!
That’ll be all for the examples. The next lesson will cover an important application of all possible selections – finding out the number of divisors of a number.