In the previous section, we saw the multiplication principle. We also saw some solved examples. In this section, we will see a few more solved examples.
Solved example 7.3
How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?
Solution:
1. Fig.7.4(a) below shows a possible combination of two numbers from the given list. The numbers are placed side by side.
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| Fig.7.4 |
• Fig.(b) shows another possible combination.
• Note that in both figs. (a) and (b), the two digit number formed, is even. This is because, the digit in the units place is even.
◼ We have to find the answer to this question:
How many such combinations are possible.
2. To try the various possible combinations, we can take two boxes as shown in fig.c.
• We have to fill the boxes with the given five digits, taking two at a time.
3. First we will fill the left side box.
• So filling of the left side box is the first event. It can take place in five different ways.
4. Next we fill the right side box. So filling the right side box is the second event.
• After filling the left box, we will be having four digits.
• But since repetition of digits is allowed, we have the option to use all the five digits for filling the right side box.
• But again, the two digit number formed must be even. So out of the five given digits, we have the option to use only two. They are: 2 and 4
• So we can write:
The second event can take place in two different ways.
5. So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 2) = 10
Solved example 7.4
Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available.
Solution:
• Five different flags are available.
◼ Note the specification in the question: at least two flags
• That means, we can make signals by
♦ putting 2 flags together
♦ putting 3 flags together
♦ putting 4 flags together
♦ putting 5 flags together
• Making a signal using a single flag is not allowed. That is why ‘at least two flags’ is specified.
1. To try various combinations using 2 flags, we can use the two boxes shown in fig.7.5(a) below:
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| Fig.7.5 |
♦ The top box can be filled in 5 different ways.
♦ The bottom box can be filled in 4 different ways
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4) = 20
2. To try various combinations using 3 flags, we can use the three boxes shown in fig.7.5(b) above.
♦ The top box can be filled in 5 different ways.
♦ The box below it can be filled in 4 different ways
♦ The bottom box can be filled in 3 different ways
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4 × 3) = 60
3. To try various combinations using 4 flags, we can use the four boxes shown in fig.7.5(c) above.
♦ The top box can be filled in 5 different ways.
♦ The box below it can be filled in 4 different ways
♦ The box below it can be filled in 3 different ways
♦ The bottom box can be filled in 2 different ways
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4 × 3 × 2) = 120
3. To try various combinations using 5 flags, we can use the five boxes shown in fig.7.5(d) above.
♦ The top box can be filled in 5 different ways.
♦ The box below it can be filled in 4 different ways
♦ The box below it can be filled in 3 different ways
♦ The box below it can be filled in 2 different ways
♦ The bottom box can be filled in 1 way
• So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 4 × 3 × 2 × 1) = 120
4. Now we can add all the possible combinations. We get:
20 + 60 + 120 + 120 = 320
Solved example 7.5
How many four digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5 if repetition of digits is not allowed?
Solution:
1. To try various combinations, we can place four boxes side by side.
2. We are given 6 digits: 0, 1, 2, 3, 4, and 5. But we cannot use ‘0’ in the first box. Because, then the number formed would be a three digit number.
• So we have 5 digits to try on the first box. That means, the first box can be filled in 5 different ways.
3. There are 6 digits available to fill in the second box. But since repetition is not allowed, we can use only 5 digits. That means, the second box can be filled in 5 different ways.
4. In a similar way,
• the third box can be filled in 4 different ways.
• the fourth box can be filled in 3 different ways.
5. So by applying multiplication principle, we get:
Total number of combinations possible = (5 × 5 × 4 × 3) = 300
The link below gives some more solved examples:
• In the next section, we will see permutations.
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