Aricent Permutation and Combinations Quiz-1

Here we need the number of possible combinations of 3 out of 5 freshmen,5C3, and the number of possible combinations of 3 out of the 15 sophomores and juniors, 15C3.

Note that we want 3 freshmen and 3 students from the other classes.

Therefore, we multiply the number of possible groups of 3 of the 5 freshmen times the number of possible groups of 3 of the 15 students from the other classes.

5C3×15C3=4,550

This second part of the problem is similar to the first except that the choice of the second group of 3 comes only from the sophomores.

Again we want 3 freshmen and 3 sophomores so we multiply the number of groups of freshmen times the number of groups of sophomores.

5C3×8C3=560

An equal number of students from each of the three classes mean 2 students from each class, that is, 2 freshmen and 2 sophomores and 2 juniors.

5C2×8C2×7C2=5,880

This part of the problem seems similar to the first three parts but it is very different.

Here we want 6 freshmen or 6 sophomores or 6 juniors.

The group cannot be all freshmen since there are only 5 freshmen.

Therefore, the group can be a group of 6 sophomores or 6 juniors.

Number of groups of sophomores =8C6=28
Number of groups of juniors=7C6=7
Since we want the number of groups of 6 sophomores or 6 juniors, we want the sum of each of these possibilities:

=28+7=35

Let the 3 boys be B1,B2,B3 and 4 prizes be P1,P2,P3 and P4.

Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)

B2 will receive prize from rest 3 available prizes(so 3 ways)

Finally, B3 will receive his prize from the rest 2 prizes available(so 2 ways)

So total ways would be: 4×3×2=24 Ways

Hence, the 4 prizes can be distributed in 24 ways.

The order of each letter in the dictionary is ABLORU.

Now, with A in the beginning, the remaining letters can be permuted in 5! ways.

Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways.

With L in the beginning, the first word will be LABORU, the second will be LABOUR.

Hence, the rank of the word LABOUR is 5!+5!+2
=242

nP3=n(n–1)(n–2)
n+1P3=(n+1)n(n–1) [Replacing n by (n+1)]

⇒ 5n(n−1)(n−2)=4(n+1)(n)(n−1)
Divide both sides by n(n−1)
5(n−2)=4(n+1)
5n−10=4n+4
5n−4n=4+10
n=14

The number of words which have at least one letter replaced:

=Total number of words −total number of words in which no letter is repeated
⇒105–16P5
⇒100000−30240
=69760

In this case the 6 students are all from the same class and we can use permutations directly.

Number of orders of freshmen =0
Number of orders of sophomores =8P6=20,160
Number of orders of juniors =7P6=5,040
Since we want the number of groups of 6 freshmen or 6 sophomores or 6 juniors, we want the sum of each of these possibilities:

0+20,160+5,040=25,200

Total number of ways in which 8064 can be resolved as the product of two factors is 24 as below:

(1,8064), (2,4032), (3,2688), (4,2016), (6,1344), (7,1152),(8,1008), (9,896), (12,672), (14,576), (16,504), (18,448), (21,884), (24,336), (28,288), (32,252), (36,224), (42,192), (48,168), (56,144), (63,128), (68,126), (72,112), (84,96).