**Question 1:** How many number greater than ten thousand can be formed from the digits 0, 1, 2, 3, 4 without repetation ?**Solution :** In order to form a number greater than 10000 we should have only 5 digits. Since, we have 5 digits we cannot take 0 in starting position.

_ _ _ _

For first digit, we have 4 choices.

For second digit, again we have 4 choices because we can include 0 from here onwards.

For third digit, we have 3 choices.

and for fourth digit only left 2 choices.

Total numbers = 4x4x3x2= 96

Hence, only 96 numbers possible.

**Question 2:** In how many ways can 4 boys and 4 girls be seated around a circular table so that no two boys are in adjacent positions?**Solution :** If we first put 4 boys around the table, we can do this in 3! ways.

Once the 4 boys are placed, we have to place 4 girls around the same table.

Now, we can see 4 vacant places are there between all 4 boys so we can do in 4! ways.

Total number of sitting arrangement = 3! x 4!

**Question 3:** Out of the 11 points in a plane, 4 are collinear. How many straight line can be formed ?**Solution :** If all points were non-collinear then possible lines would have been ^{11}C_{2}.

But, 4 points are collinear lie on the same line. So, they are all counted as a single line.

Total number of straight line = ^{11}C_{2} – ^{4}C_{2} + 1

= 11×10/2 – 4×3/2 + 1

= 55 – 6 + 1

= 50

**Question 4:** Twenty people attend a party and shake hands with one another.In how many ways hand shake is possible?**Solution :** All people shake hands with one another except himself.

1st person has 19 hand shakes and 2nd also has 19 hand shakes ………and so on

20 people x 19 hand shakes

and we know A handshake with B or B handshake with A, it is counted as 1 handshake.

So, total number of handshake = 20×19/2 = 190**Shortcut :**

for n people there are always ^{n}C_{2} handshakes.

**Question 5:** How many different sums of money can be formed from the four type of notes Rs 10, Rs 20, Rs 50 and Rs 100 ?**Solution :** Type of notes = 4

So, total number of sum can be formed = 2^{4} – 1 = 15

**Question 6:** Five chocolates of different flavours are to be distributed in three different children such that any child get at least 1 chocolate. What is the maximum number of different ways in which this can be distributed?**Solution :** Acc. to question

Chocolates can be distributed as [(3, 1, 1)(1, 3, 1)(1, 1, 3)]

or [(2, 2, 1)(2, 1, 2)(1, 2, 2)]

Total number of ways = 3 x ^{5}C_{3} x ^{2}C_{1} x ^{1}C_{1} + 3 x ^{5}C_{2} x ^{3}C_{2} x ^{1}C_{1}

= 60 + 90

= 150

**Question 7:** Ram and his wife Sita both have five friends each. Ram has 2 boys and 3 girls. Sita has 3 boys and 2 girls. In how many maximum number of different ways can they invite 2 boys and 2 girls such that two of them are Ram’s friend and two are Sita’s friend?**Solution :** Selection can be done like that

i) 2 boys from Ram’s friends and 2 girls from his wife Sita’s friends OR

ii) 1 boy and 1 girl from Ram’s friends and 1 boy and 1 girl from Sita’s friends OR

iii) 2 boys from his wife Sita’s friends and 2 girls from Ram’s friends

Total number of ways = ^{2}C_{2}.^{2}C_{2} + ^{3}C_{1} x ^{2}C_{1} x ^{3}C_{1} x ^{2}C_{1} + ^{3}C_{2}.^{3}C_{2}

= 1 + 36 + 9

= 46 ways

**Question 8:** In how many ways can 4 notebooks can be distributed to 5 students if each can get any number of notebooks?**Solution :** Since all the notebooks are identical or distinct we don’t know.

So, we take all are distinct and it can be distributed in 5^{4} ways.

**Question 9:** How many batting orders are possible for the Indian cricket team if there is a squad of 16 to choose from such that Virat Kohli and Rohit Sharma are always chosen?**Solution :** We need to select 9 players out of 14 players beacause two of them is already selected.

The selection of 11 players can be done in^{14}C_{9} ways.

But batting order is also required to calculate for these 11 players so arrangements can be done in 11! ways.

Total number of batting orders possible = ^{14}C_{9} . 11!

**Question 10:** How many motor vehicle registration numbers plates can be formed for the state Haryana having code like (HR 12Q 8702) with the digits 0, 1, 2, 3, 4, 5, 6 and contains consonant at the alphabetical place(No digits being repeated).**Solution :** For Haryana state number plate always contains HR in starting.

Two consonants already used in HR so remaining consonant = 21 -2 = 19.

So, possible number of plates = 1(HR)x7x6x19(consonants)x5x4x3x2 = 95760

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