How many words of 4 letters with or without meaning be made from the letters of the word leading when repetition of letters is?

Exercise :: Permutation and Combination - General Questions

13.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A.	10080
B.	4989600
C.	120960
D.	None of these

Answer: Option C

Explanation:

In the word 'MATHEMATICS', we treat the 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 and the rest are different.

Number of ways of arranging these letters =	8!	= 10080.
(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =	4!	= 12.
2!

Required number of words = (10080 x 12) = 120960.

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Exercise :: Permutation and Combination - General Questions

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer: Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

Text Solution

Answer : 7

Solution : There are 10 letters in the word LOGARITHMS. So, the number of 4-letter words is equal to the number of arrangements of 10 letters, taken 4 at a time, i.e., `.^(10)P_(4)=5040`.

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In how many ways 5 rings can be worn on 3 fingers? A) 15 B) 120 C) 60 D) 70 E) 243

C) 60
Explanation: 0 0 0 Let these 3 circles are 3 fingers For 1st finger we have 5 choices, for second finger we have 4 choices left of rings, for third finger we have 3 choices left. So total 5*4*3 = 60 ways
In how many ways the letters of the word ‘AUTHOR’ be arranged taking all the letters? A) 120 B) 720 C) 360 D) 60 E) None of these

B) 720
Explanation: AUTHOR contains 6 letters, so total 6! ways.
In how many ways the letters of the word ‘MINIMUM’ be arranged taking all the letters? A) 420 B) 840 C) 5040 D) 720 E) 360

A) 420
Explanation: MINIMUM contains 7 letters, so total 7! ways. But it contains 2 I’s and 3 M’s so divide by 2! And 3! So ways 7!/(2! * 3!) = 7*6*5*4*3*2*1 / 2*1*3*2*1 = 420
How many words of 4 letters with or without meaning be made from the letters of the word ‘LEADING’, when repetition of letters is allowed? A) 4808 B) 57600 C) 2401 D) 57624 E) None of these

D) 57624
Explanation: LEADING is 7 letters. We have 4 places where letters are to be placed. For first letter there are 7 choices, since repetition is allowed, for second, third and fourth letter also we have 7 choices each, so total of 7*7*7*7 ways = 2401 ways. Now for arrangement of these 4 words, we have 4! Ways. So total of 2401 * 4! Ways.
In how many ways letters of word ‘INVISIBLE’ be arranged such that all vowels are together? A) 2560 B) 2880 C) 5040 D) 2520 E) 720

B) 2880
Explanation:
First make IIIE in a circle. So we have
Now we have N, V, S, B, L and box, their arrangements can be done in 6! Letters inside circle are also to be arranged, we have I, I, I, E so ways are 4!/3! Total ways 6! * 4!/3!
How many words can be made out of the letters of word ‘POUNDING’ such that all vowels occupy odd places? A) 1440 B) 1400 C) 7200 D) 5600 E) 40320

A) 1440
Explanation: In POUNDING, there are 8 places 1 2 3 4 5 6 7 8
So for 3 places selection of vowels, we have 1, 3, 5, 7 number places 4C3 ways
Now for arranging these 3 vowels, ways are 3! Remaining 5 are consonants (in which there are 2 N’s) for which 5!/2!
so total ways = 4C3*3!*(5!/2!)
In how many ways a group of 2 men and 4 women be made out of a total of 4 men and 7 women? A) 720 B) 210 C) 420 D) 360 E) 120

B) 210
Explanation: We have to select 2 men from 4 men, and 4 women from 7 women
So total ways = 4C2*7C4
There are 8 men and 7 women. In how many ways a group of 5 people can be made such that at least 3 men are there in the group? A) 1545 B) 1626 C) 1722 D) 1768 E) 1844

C) 1722
Explanation: Case 1: 3 men and 2 women
8C3*7C2 = 1176
Case 2: 4 men and 1 women
8C4*7C1 = 490
Case 3: all 5 men
8C5 = 56

Add all the cases.
There are 6 men and 7 women. In how many ways a committee of 4 members can be made such that a particular woman is always included. A) 180 B) 120 C) 240 D) 220 E) 260

D) 220
Explanation:
There are total 13 people, a particular woman is to be included, so now 12 people are left to chosen from and 3 members to be chosen. So ways are 12C3.
There are 5 men and 3 women. In how many ways a committee of 3 members can be made such that 2 particular men are always to be excluded. A) 50 B) 20 C) 24 D) 48 E) None of these

B) 20
Explanation:
Total 8 people, 2 men are to excluded, so 6 men left to be chosen from and 3 members to be chosen. So ways are 6C3.