WeCP Permutation and Combination Quiz 1

Explanation :

A team of 6 members has to be selected from the 10 players. This can

be done in ^{20}C_{6} ways  = 38760

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 38760×6= 232,560

Explanation :

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as in the dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

Explanation :

There are 100 people in the room ,so our n value is 100

No. of handshakes   = ^{100}C_{2}

= 4950

Explanation :

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

Explanation :

From 7 sharpeners 4 sharpeners can be selected as ^{7}C_{4} ways

From 5 erasers 3 erasers can be selected in ^{5}C_{3} ways.

Now with every selection, the number of ways of arranging 7 objects is ^{7}P_{7} ways.

Total numbers of sets = ^{7}C_{4} \times ^{5}C_{3} \times ^{7}P_{7}

= 35 x 10 x 5040 =1764000

Explanation :

Let 7 coca-cola be one unit and now there are 7 units in all.

They can be arranged in 7! ways.

In each of these arrangements 7 coca-cola can be arranged in 7! ways.

Total number of arrangements in which coca-cola are always together = 7! x 7!= 5040 x 5040 = 25401,600

Explanation :

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = ^{7}C_{3} = 35

Explanation :

Let the number of sides be n.

The number of diagonals is given by ^{n}C_{2 - n}

Therefore, ^{n}C_{2 - n} = 44, n > 0

^{n}C_{2 - n} = 44

n^{2} - 3n - 88 = 0

n^{2} -11n + 8n - 88 = 0 

n(n - 11) + 8(n - 11) = 0

n = -8 or n = 11.

As n>0, n will not be -8. Therefore, n=11.

Explanation :

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! Ways.

Explanation :

There are 15 dots in total,and to make a triangle we need to select any three of those dots.

^{15}C_{3} = 455