How many ways can a team of 5 members be formed by 6 boys and 4 girls if atleast 2 girls should be selected?

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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.

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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.

Out of 6 boys and 4 girls, a committee of 5 is to be formed. In how many ways can this be done if at least two girls are to be included in the committee. 2

Text Solution

Solution : (a) We have to make a selection of (i) (2 ladies out of 4) and (3 men out of 6) or (ii) (3 ladies out of 4 ) and (2 men out of 6) or (iii) (4 ladies out of 4) and (1 man out of 6). The number of ways of these selections are: Case I `.^(4)C_(2)xx.^(6)C_(3)=6xx20=120`. Case II `.^(4)C_(3)xx.^(6)C_(2)=4xx15=60`. Case III `.^(4)C_(4)xx.^(6)C_(1)=1xx6=6`. Hence, the required number of ways =(120+60+6)=186. (b) We have to make a selection of (i) (1 lady out of 4) and (4 men out of 6) or (ii) (2 ladies out of 4) and (3 men out of 6). The number of ways of these selections are: Case I `.^(4)C_(1)xx.^(6)C_(4)=4xx15=60`. CaseII `.^(4)C_(2)xx.^(6)C_(4)=4xx15=60`. Case II `.^(4)C_(2)xx.^(6)C_(3)=6xx20=120`. Hence, the required number of ways = (60+120)=180.