Combination Function and its properties MCQ Quiz - Objective Question with Answer for Combination Function and its properties - Download Free PDF

Explanation:

Since there are 4 lines, the maximum number of intersection points are ⁴C₂ = 6.

The maximum number of intersections of the circle with 4 lines is 4 × 2 = 8.

Required intersection points = 6 + 8 = 14

∴ the correct answer is Option 3 ie 14

Calculation

Suppose the two players did not play at all so that the remaining (n−2)" id="MathJax-Element-19452-Frame" role="presentation" style="position: relative;" tabindex="0">(n−2)(n−2) $\left(n-2\right)$ players played \(^{n-2}C_{2}\) matches.

Since these two players played 3 matches each, hence the total number of matches is \(^{n-2}C_{2}\) + 3 + 3 = 84 (given).

\(\dfrac{(n-2)(n-3)}{2} = 78\)

or \(n^{2} - 5n + 6 = 156\) or \(n^{2} - 5n - 150 = 0\)

or \((n-15)(n+10) = 0\) \(\therefore n = 15\) \((n \neq -10)\)

\(2w, xM\)

\(2\) \(^{x}{C_2} - 2 \times 2x = 66\)

\(^{x}{C_2} - 2x = 33\)

\(x(x-1) - 4x = 66\)

\(x = 11\)

Calculation

Given

⁶C_m + 2(6Cm+1) + 6Cm+2  > 8C3

⇒ 7Cm+1 + 7Cm+2  > 8C3

⇒ 8Cm+2  > 8C3

⇒ m = 2

And ^n-1P₃ : ⁿP₄ = 1 : 8 

⇒ \( \frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{8}\)

⇒ n = 8 
∴ⁿP_m+1 + ⁿ⁺¹C_m = ⁸P₃ + ⁹C₂

⇒ _{\( 8 \times 7 \times 6+\frac{9 \times 8}{2} \)}

⇒ 372

Hence, Option (4) is correct

Concept:

The formula for n factorial is:

n! = n × (n−1)!

This means that the factorial of any number is, the given number, multiplied by the factorial of the previous number.

Calculation:

According to the question

The arrangement be like B G B G B G B

Now, we can see here

4 Boys can be seated in 4! Ways

And 3 Girls can be seated in 3! Ways

So, The required number of ways = 4! × 3!

⇒ 144

∴ The required number of ways is 144.

Concept:

• ⁿC_r = ⁿC_n-r.
• If nCx = nCy, then x = y or x + y = n.
• nCr + nCr-1 = n+1Cr.

Calculations:

\(\rm \begin{pmatrix} 15\\ 8\end{pmatrix}+\begin{pmatrix} 15\\ 7\end{pmatrix}=\begin{pmatrix} \rm n\\ \rm r\end{pmatrix}\)

⇒ 15C8 + 15C7 = nCr

Using nCr + nCr-1 = n+1Cr:

⇒ ​15C8 + 15C8-1 = ¹⁵⁺¹C₈ = nCr

⇒ ​16C8 = nCr

⇒ n = 16 and r = 8.

Concept:

The notation C(n, r) is the number of combinations/groups of n different things taking r at a time and is given by: \(C\;\left( {n,\;r} \right) = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n\;}} - {\rm{\;r}}} \right)!}}\)

If C(n, x) = C(n, y), then x + y = n.

Calculation:

Given C(20, n + 2) = C(20, n – 2)

As we know that, if C(n, x) = C(n, y), then x + y = n.

⇒ (n + 2) + (n – 2) = 20

⇒ 2n = 20

Concept:

If \(\rm ^nC_x={^nC_y}\) then x + y = n

\(\rm ^nC_r=\frac{n\times(n-1)\times....(n-r+1)}{r!}\)

Calculation: 

Given: nC5 = nC7

It is possible only when n = 5 + 7 = 12

∴ \(\rm ^nC_4=^{12}C_4=\frac{12\times11\times10\times9}{4!}\)

\(=\frac{12\times11\times10\times9}{4\times3\times2}\)

= 495

Hence, option (3) is correct.

Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: \({\;^n}{C_r} = \frac{{n!}}{{r!\; \times \left( {n - r} \right)!}}\)

Calculation:

Given: There are 5 subjects and for a student to pass in an examination the students has to pass in each of the 5 subjects.

Here, we have to find in how many ways a student can fail in the examination.

In order to fail in the examination the student can fail in 1 or 2 or 3 or 4 or 5 subjects out of the 5 subjects in each case.

Case 1: Student fails in any 1 subject out of the 5 subjects.

No. of ways in which student can fail in any 1 subject out of the 5 subjects = \({\;^5}{C_1}\)

Case 2: Student fails in any 2 subjects out of the 5 subjects.

No. of ways in which student can fail in any 2 subjects out of the 5 subjects = \({\;^5}{C_2}\)

Case 3: Student fails in any 3 subjects out of the 5 subjects.

No. of ways in which student can fail in any 3 subjects out of the 5 subjects = \({\;^5}{C_3}\)

Case 4: Student fails in any 4 subjects out of the 5 subjects.

No. of ways in which student can fail in any 4 subjects out of the 5 subjects = \({\;^5}{C_4}\)

Case 5: Student fails in all 5 subjects.

No. of ways in which student can fail all 5 subjects. = \({\;^5}{C_5}\)

∴ Total number of ways in which a student can fail in an examination = \({\;^5}{C_1}\) + \({\;^5}{C_2}\) + \({\;^5}{C_3}\) + \({\;^5}{C_4}\) + \({\;^5}{C_5}\) = 31

Concept:

The number of all combinations of n distinct things taken r at a time (r ≤ n) is:

\(^{n}C_{r}=\frac{n!}{(n-r)!r!}\)

Calculation:

Since 2nC3 : nC2 = 12 : 1

\(⇒ \frac{2n!}{(2n-3)!\space 3!}:\frac{n!}{(n-2)!\space 2!}=12:1\)

\(⇒ \frac{2n(2n-1)(2n-2)}{3\times 2\times 1}:\frac{n(n-1)}{2\times 1}=12:1\)

⇒ 2n - 1 = 9

⇒ n = 5

Concept:

• nCr = nCn - r.
• If nCx = nCy, then x + y = n.
• nCr + nCr - 1 = n + 1Cr.

Calculation:

Given: nC15 = nC8

As we know, nCx = nCy, then x + y = n.

So, n = 15 + 8 = 23

Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: \({^n}{C_r} = \frac{{n!}}{{r!\; × \left( {n - r} \right)!}}\)

Calculation:

Given: There are 7 consonants and 4 vowels

Here, we have to find how many word can be formed such that it has 3 consonants and 2 vowels.

No. of ways to select 2 vowels out of 4 vowels = \({^4}{C_2} \)

No. of ways to select 3 consonants out of 7 consonants = \({^7}{C_3} \)

∴ No. of words that can be formed which contains 3 consonants and 2 vowels = \({^4}{C_2} \) × \({^7}{C_3} \)

As we know that, \({^n}{C_r} = \frac{{n!}}{{r!\; × \left( {n - r} \right)!}}\)

⇒ \({^4}{C_2}\) × \({\;^7}{C_3} \)  = 6 × 35 = 210

The no. of ways to arrange words containing 3 consonants and 2 vowels = 210 × 5! = 25200

Hence, option 3 is the correct answer

Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: \({\;^n}{C_r} = \frac{{n!}}{{r!\; \times \left( {n - r} \right)!}}\)

Calculation:

Given: The group has 10 men and 8 women.

Here, we need to form a team of 3 men or 2 women .

As we know that, the  number of ways to select r things out of n given things wherein r ≤ n is given by: \({\;^n}{C_r} = \frac{{n!}}{{r!\; \times \left( {n - r} \right)!}}\)

∴ No. of ways to form a team of 3 men \(= {\;^{10}}{C_3} = \frac{{10!}}{{3!\; \times \left( {10 - 3} \right)!}}\)

∴ No. of ways to form a team of 2 women \(= {\;^8}{C_2} = \frac{{8!}}{{2!\; \times \left( {8 - 2} \right)!}}\)

∴No. of ways to form a team of 3 men or 2 women \(= {\;^{10}}{C_3} + {\;^{8}}{C_2} = 148\)

Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: \({\;^n}{C_r} = \frac{{n!}}{{r!\; × \left( {n - r} \right)!}}\)

Calculation:

Given: There are 6 teachers and 8 students  out of which we need to form a committee of 11 members such that the committee has exactly 4 teachers in it.

If out 11 members of committee 4 are teachers then 7 members are students.

So, the committee compromises of 4 teachers and 7 students.

No. of ways in which 4 teachers can be selected from 6 teachers = \({\;^{6}}{C_4}\)

No. of ways in which 7 students can be selected from 8 students = \({\;^{8}}{C_7}\)

So, the number of ways in which the committee can be formed = \({\;^{6}}{C_4} \times {\;^{8}}{C_7}\)

As we know that, \({\;^n}{C_r} = \frac{{n!}}{{r!\; × \left( {n - r} \right)!}}\)

⇒ \({\;^{6}}{C_4} \times {\;^{8}}{C_7} = 120\)

Hence, the committee can be formed in 120 ways

Concept:

Combinatorics:

• ⁿC_r = \(\rm \frac {n!}{r!(n\ -\ r)!}\).
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

Calculation:

Given equation is: 3(6C3) = 10(xC2)

⇒ \(\rm 3×\left[\frac {6!}{3!(6\ -\ 3)!}\right]=10×\left[\frac {x!}{2!(x\ -\ 2)!}\right]\)

⇒ \(\rm 3×\left(\frac {6\ ×\ 5\ ×\ 4}{3\ ×\ 2\ ×\ 1}\right)=10×\left[\frac {x(x\ -\ 1)}{2\ ×\ 1}\right]\)

⇒ x(x - 1) = 12 = 4 × 3

⇒ x = 4.