PE 2 - Permutation & Combination | Modern Math - Permutation & Combination

Answer: 3186

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Explanation :

The total number of ways in which 7 caps can be put on 7 different markers is 7! = 5040.

Number of ways of putting caps on markers such that cap and body colour does not match for any marker = $7!\times \left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}\right]$ = 2520 – 840 + 210 – 42 + 7 – 1 = 1854,

∴ Number of ways in which the caps can be put on the body of the markers such that colour of the cap and the body for at least 1 marker is the same = 5040 – 1854 = 3186.

Hence, 3186.

Answer: Option E

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Explanation :

We have to form a 5-digit number using digits 0, 2, 4, 6 and 8.

The first digit can be filled in 4 ways i.e., 2 or 4 or 6 or 8
The remaining digits can be filled in 5 ways i.e., 0 or 2 or 4 or 6 or 8

∴ Total number of 5-digit numbers with all its digits as even = 4 × 5 × 5 × 5 × 5 = 2500

Hence, option (e).

Answer: Option A

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Explanation :

As all the books are from different publishers, they are all different. 

∴ Required number of ways to arrange the books = Total number of ways to arrange (4 + 3 + 2 =) 9 different books on a shelf = 9! = 362880

Hence, option (a).

Answer: Option D

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Explanation :

Considering the 4 girls who want to sit together to be a single person, there will be 7 persons now. They can be arranged in 6! ways. 

The 4 girls can be arranged among themselves in 4! ways.

∴ Required number of ways = 6! × 4!

Hence, option (d).

Answer: 1560

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Explanation :

Since all rooms must be occupied, this can happen in two ways. 3, 1, 1, 1 or 2, 2, 1, 1.

Case 1: 3, 1, 1, 1 i.e., out 6 friends 3 will stay together and other 3 will stay individually in different rooms.

The 3 friends who stay together can se selected in ⁶C₃ = 20 ways.

Now, we have 4 different groups of friends (i.e., 3, 1, 1, 1) who need to stay in 4 different rooms. This can be done in 4! = 24 ways.

∴ Total number of ways = 20 × 24 = 480.

Case 2: 2, 2, 1, 1 i.e., out 6 friends 2 will stay together, another pair of 2 will stary in other rooms and 2 remaining friends will stay individually in different rooms.

The 2 groups of 2 friends each who stay together can se selected in ⁶C₂ × ⁴C₂ ÷ 2 = 45 ways.

Now, we have 4 different groups of friends (i.e., 2, 2, 1, 1) who need to stay in 4 different rooms. This can be done in 4! = 24 ways.

∴ Total number of ways = 45 × 24 = 1080.

⇒ Total number of ways 6 friends can stay in 4 different rooms = 480 + 1080 = 1560.

Hence, 1560.

Answer: Option C

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Explanation :

We know:

(a + b)n = ⁿC₀ × a⁰ × bⁿ + ⁿC₁ × a¹ × b^n-1 + ⁿC₂ × a² × b^n-2 + … + ⁿC_n × aⁿ × b⁰ 

Let the term independent of x in ${\left[x^2-\frac{1}{x}\right]}^{12}$ is ⁿC_k × (x²)^k × ${\left(\frac{1}{x}\right)}^{n-k}$

= ⁿC_k × x^2k × x^k-n

= ⁿC_k × x^3k–n

Now for this term to be independent of x, 3k – n = 0

⇒ k = n/3 = 12/3 = 4

∴ The term becomes ¹²C₄.

Hence, option (c).

Answer: Option C

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Explanation :

We need to select 4 people out of 5 couples such that no couple is selected i.e., at max only one person will be selected from a couple.

First, we select 4 couples and then we will pick one from each of these 4 couples.

Number of ways of selecting 4 couples = ⁵C₄ = 5 ways.

From each of these 4 couples either the female or the male could be selected i.e., 2 ways of selection.

∴ Total number of ways of selecting 4 people = 5 × (2)⁴ = 80.

Hence, option (c).

Answer: Option A

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Explanation :

Case 1: Indian wins all 5 matches
This can happen in only 1 way.

Case 2: Indian wins 4 matches and loses 1 match i.e., WWWWL
Number of ways = 5!/4! = 5 ways

Case 3: Indian wins 3 matches and loses 2 matches i.e., WWWLL
Number of ways = 5!/3!2! = 10 ways

∴ Total number of ways = 1 + 5 + 10 = 16 ways

Hence, option (a).

Answer: Option C

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Explanation :

Case 1: Indian wins 4 matches and draws 1 match i.e., WWWWD
Number of ways = 5!/4! = 5 ways

Case 2: Indian wins 3 matches, loses 1 match and draws 1 match i.e., WWWLD
Number of ways = 5!/3! = 20 ways

∴ Total number of ways = 5 + 20 = 25 ways

Hence, option (c).

Answer: Option B

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Explanation :

Case 1: Indian wins all 5 matches i.e., WWWWW
This can happen in only 1 way.

Case 2 (a): Indian wins 4 matches and loses 1 match i.e., WWWWL
Number of ways = 5!/4! = 5 ways

Case 2 (b): Indian wins 4 matches and draws 1 match i.e., WWWWD
Number of ways = 5!/4! = 5 ways

Case 3 (a): Indian wins 3 matches and loses 2 matches i.e., WWWLL
Number of ways = 5!/3!2! = 10 ways

Case 3 (b): Indian wins 3 matches, loses 1 match and draws 1 match i.e., WWWLD
Number of ways = 5!/3! = 20 ways

Case 3 (c): Indian wins 3 matches and draws 2 matches i.e., WWWDD
Number of ways = 5!/3!2! = 10 ways

Case 4 (a): Indian wins 2 matches, loses 1 match and draws 2 matches i.e., WWLDD
Number of ways = 5!/2!2! = 30 ways

Case 4 (b): Indian wins 2 matches, and draws 3 matches i.e., WWDDD
Number of ways = 5!/2!3! = 10 ways

Case 5: Indian wins 1 match, and draws 4 matches i.e., WDDDD
Number of ways = 5!/4! = 5 ways

∴ Total number of ways = 1 + 5 + 5 + 10 + 20 + 10 + 30 + 10 + 5 = 96 ways

Hence, option (b).