1.Evaluate 30!/28!
Sol:- 30!/28! = 30 * 29 * (28!) / (28!)
= 30 * 29 =870
2.Find the value of 60P3
Sol:- 60P3 = 60! / (60 - 3)! = 60! / 57!
= (60 * 59 *58 * (57!) )/ 57!
= 60 * 59 *58
= 205320
3. Find the value of 100C98 50C 50
Sol:- 100C98 = 100C100 - 98)
= 100 * 99 / 2 *1
= 4950
50C50 = 1
4.How many words can be formed by using all the letters of the word "DAUGHTER" so that vowels always come together & vowels are never
together?
Sol:- (i) Given word contains 8 different letters
When the vowels AUE are always together we may suppose
them to form an entity ,treated as one letter
then the letter to be arranged are DAHTR(AUE)
these 6 letters can be arranged in 6p6 = 6!
= 720 ways
The vowels in the group (AUE) may be arranged in 3! = 6 ways
Required number of words = 760 * 6 =4320
(ii) Total number of words formed by using all the letters of the given
words
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 40320
Number of words each having vowels together
= 760 * 6
= 4320
Number of words each having vowels never together
= 40320 - 4320
= 36000
5.In how many ways can a cricket eleven be chosen out of a batch
of 15 players.
Sol:- Required number of ways
= 15C 11 = 15C (15-11)
= 15 C 4
15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1
= 1365
6.In how many a committee of 5 members can be selected from 6 men
5 ladies consisting of 3 men and 2 ladies
Sol:- (3 men out of 6) and (2 ladies out of 5) are to be chosen
Required number of ways
=(6C3 * 5C2)
= 200
7.How many 4-letter word with or without meaning can be formed out
of the letters of the word 'LOGARITHMS' if repetition of letters is
not allowed
Sol:- 'LOGARITHMS' contains 10 different letters
Required number of words
= Number of arrangements of 100 letters taking
4 at a time
= 10P4
= 10 * 9 * 8 * 7
= 5040
8.In how many ways can the letter of word 'LEADER' be arranged
Sol:- The word 'LEADER' contains 6 letters namely
1L, 2E, 1A, 1D and 1R
Required number of ways
= 6! / (1!)(2!)(1!)(1!)(1!)
= 6 * 5 * 4 * 3 * 2 *1 / 2 * 1
=360
9.How many arrangements can be made out of the letters of the word
'MATHEMATICS' be arranged so that the vowels always come
together
Sol:- In the word 'MATHEMATICS' we treat vowels
AEAI as one letter thus we have MTHMTCS(AEAI)
now we have to arrange 8 letters out of which M occurs
twice ,T occurs twice & the rest are different
Number of ways of arranging these letters
= 8! / (2!)(2!)
= 10080
now AEAI has 4 letters in which A occurs 2 times and the rest
are different
Number of ways of arranging these letters
= 4! / 2! = 12
Required number of words = (10080 * 12)
