Topic - Permutation & Combination

Solved ? - Yes

Source - 2IIM ]]>

Topic - Permutation & Combination

Solved ? - Yes

Source - 2IIM ]]>

Rearrangements where the A’s appear together:

Put 4 A’s together into X and rearrange MLYLMX. This can be rearranged in 6!/2!2!

Rearrangements where the consonants appear together: Put consonants together into Z. The word will be AAAAZ. This can be rearranged in 5!/4! ways.

Now, Z is MMLLY. Z can take 5!/2!2! ways.

So, total number of arrangements =5!/4! × 5!/2!2! ]]>

So, we need to see how many times the digit 4 gets printed from 1 to 2032 in base 8.

Let us consider a number of the form

When c = 4, a can take all values from 0 to 7 and b can take all values from 0 to 7. So, 4 gets printed at c 64 times.

When b = 4, a can take all values from 0 to 7 and c can take all values from 0 to 7. So, 4 gets printed at b 64 times.

When a = 4, b can take all values from 0 to 7 and c can take all values from 0 to 7. So, 4 gets printed at a 64 times.

So, totally 4 gets printed 64 × 3 = 192 times from

From

So, up to

We need to account for all base 8 numbers from

In these 26 numbers, the digit 4 gets printed thrice

So, the digit 4 gets printed 384 + 3 = 387 times. ]]>

This can happen in 2 + 2 = 4 ways. ]]>

Scenario II: 5 distinct consonants and 1 vowel appearing twice – Number of words = 21C5 × 5C1 × 7!/2! ]]>

Number of ways = 6C4 + 6C5 + 6C6

= 15 + 6 + 1

= 22 ]]>

Scenario II: 4 face cards and 0 numbered cards: 12C4

Therefore, total number of ways is 12C3 × 36C1 + 12C4 ]]>

There are 3 different CDs. Each CD can be either picked or not picked. So, total number of options =

Of these there is one option where no CD is picked. We need to exclude that option.

So, number of possibilities =

Total number of outcomes = 4 × (

= 4 × 7

= 28

So, we need to add these 12 numbers.

Let us consider the units’ digit of these 12 numbers.

The units digit will be the one of the digits 2, 3, or 5. If the last digit were 3, the first 3 digits should be some rearrangement of 2, 2, and 5. So, there are 3!/2! Such numbers. Or, 3 such numbers.

Similarly there are three numbers with 5 as the units digit.

If the last digit were 2, the first 3 digits should be some rearrangement of 2, 3, and 5. So, there are 3! such numbers, or, 6 such numbers.

So, the units digit will be 2 for six numbers, 3 for three numbers, and 5 for three numbers. Sum of all these unit digits will be 2 × 6 + 3 × 3 + 5 × 3 = 12 + 9 + 15 = 36.

Sum of all the tens digits will be 36. Sum of all the digits in the hundreds’ place will be 36.

Sum of all the digits in the thousands’ place is 36.

So, sum of all the 4–digit numbers will be 36 × 1111 = 39996. ]]>

Sum of the digits = 2: This can only be 1 + 1. One way

Sum of the digits = 3: This can be 1 + 2 or 2 + 1. 2 ways

Sum of the digits = 4: 1 + 3, 3 + 1, 2 + 2. 3 ways

Sum of the digits = 5: 1 + 4, 4 + 1, 3 + 2, 2 + 3. 4 ways

Sum of the digits = 6: 1 + 5, 5 + 1, 2 + 4, 4 + 2, 3 + 3. 5 ways

Sum of the digits = 7: 1 + 6, 6 + 1, 2 + 5, 5 + 2, 3 + 4, 4 + 3. 6 ways

Sum of the numbers in the two throws can be less than 8 in 1 + 2 + 3 + 4 + 5 + 6 = 21 ways.

We notice a very simple pattern here. Try the sum of the numbers all the way to 12 and see the rest of the pattern also. ]]>

For choosing any subset, each element can either be part of the subset or not part of the subset. So, for each element, there are two options. So, with 10 elements in the set, we can create

Subsets whose product is not a multiple of 3 = Subsets of the set {1, 2, 4, 5, 7, 8, 10} =

3 engineers and 2 doctors: 4C3 × 5C2 = 4 × 10 = 40

4 engineers and 1 doctor : 4C4 × 5C1 = 1 × 5 = 5

Total number of possibilities = 40 + 5 = 45 ]]>

For a given set of four cards, say 2345, they can be from 4 different suits in 4! ways.

So, total number of possibilities = 11 × 4! = 264. ]]>