JEE Advanced (Single Correct MCQs): Permutations and Combinations - JEE MCQ

ⁿC_{r -1} = 36, ⁿC_r = 84,ⁿC_r+1= 126
KEY CONCEPT : We know that

⇒    ....(1)

⇒ 

⇒ 2n – 5r – 3 = 0 ....(2)

Solving (1) and (2), we get n = 9 and r = 3.

Total number of words that can be formed using 5 letters out of 10 given different letters
= 10 × 10 × 10 × 10 × 10 (as letters can repeat)
= 1, 00, 000
Number of words that can be formed using 5 different letters out of 10 different letters
= ¹⁰P₅ (none can repeat)

∴ Number of words in which at least one letter is repeated
= total words–words with none of the letters repeated
= 1,00,000 – 30,240  
= 69760

Two women can choose two chairs out of 1, 2, 3, 4, in ⁴C₂ ways and can arrange themselves in 2! ways.
Three men can choose 3 chairs out of 6 remaining chairs in ⁶C₃ ways and can arrange themselves in 3! ways
∴ Total number of possible arrangements are
⁴C₂× 2! × ⁶C₃ × 3!=⁴P₂× ⁶P₃

KEY CONCEPT : We know that a number is divisible by 3 if the sum of its digits is divisibly by 3.
Now out of  0, 1, 2, 3, 4, 5 if we take 1, 2, 3, 4, 5 or 0, 1, 2, 4, 5 then the 5digit numbers will be divisible by 3.
Case I : Number of 5 digit numbers formed using the
digits 1, 2, 3, 4, 5 = 5! = 120
Case II : Taking 0, 1, 2, 4, 5 if we make 5 digit number then
I place can be filled in = 4 ways (0 can not come at I place)
II place can be filled in = 4 ways
III place can be filled in = 3 ways
IV place can be filled in = 2 ways
V place can be filled in = 1 ways
∴ Total numbers are = 4 × 4! = 96
Thus total numbers divisible by 3 are
= 120 + 96 = 216

X – X – X – X – X. The four digits 3, 3, 5, 5 can be arranged at (–) places in  = 6 ways.

The five digits 2, 2, 8, 8, 8 can be arranged at

(X) places in  = 10ways.

Total no. of arrangements = 6 × 10 = 60 ways.

As per question,

⇒ n(n - 1) (n + 1 -n + 2)= 126
⇒ n (n – 1) = 42 ⇒ n (n – 1) = 7 × 6  ⇒ n = 7.

Total number of ways of arranging the letters of the word BANANA is = 60 Number of words in which 2 N’s come together is = 20

Hence the required number = 60 – 20  = 40

If we see the blocks in terms of lines then there are 2m vertical lines and 2n horizontal lines. To form the required rectangle we must select two horizontal lines, one even numbered (out of 2, 4, .....2n) and one odd numbered (out of 1, 3....2n–1) and similarly two vertical lines. The number of rectangles is
^mC₁ .  ^mC₁ . ⁿC₁ . ⁿC₁ = ^m2n₂

∵ r, s, t are prime numbers,
∴  Section of (p, q) can be done as follows

∴ r can be selected 1 + 1 + 3 = 5 ways

Similarly s and t can be selected in 9 and 5 ways respectivley.
∴ Total ways = 5 × 9 × 5 = 225

The letter of word COCHIN in alphabetic order are C, C, H, I, N, O.
Fixing first letter C and keeping C at second place, rest 4 can be arranged in 4! ways.
Similarly the words starting with CH, CI, CN are 4! in each case.
Then fixing first two letters as CO next four places when filled in alphabetic order give  the word COCHIN.
∴ Numbers of words coming before COCHIN are
4 × 4! = 4 × 24 = 96

We h ave to for m 7 digit n umber s, using th e digits 1, 2 and 3 only, such that the sum of the digits in a number = 10.
This can be done by taking 2, 2, 2, 1, 1, 1, 1, or by taking 2, 3, 1, 1, 1, 1, 1.
∴ Number of ways

∵ Each person gets at least one ball.
∴ 3 Per son s can have 5 balls in th e followin g systems

or

The number of ways to distribute the balls in first system =  
⁵C₁ x ⁴C₁ x ³C₃

Also 3, persons having 1, 1 and 3 balls can be arranged in   ways.

∴ No. of ways to distribute 1, 1, 3 balls to the three persons

⁵C₁ x ⁴C₁ x ³C₃ x

Similarly the total no. of ways to distribute 1, 2, 2 balls to the three persons = ⁵C₁ x ⁴C₂ x ²C₂ x

∴ The required number of ways = 60 + 90 = 150

∵ Car d numbered 1 is always placed in envelope numbered 2, we can consider two cases.
Case I: Card numbered 2 is placed in envelope numbered 1.
Then it is dearrangement of 4 objects, which can be done in

 = 9 ways

Case II: Card numbered 2 is not placed in envelope numbered 1.
Then it is dearrangement of 5 objects, which can be done in

= 44 ways

∴ Total ways = 44 + 9 = 53

Either one boy will be selected or no boy will be selected. Also out of four members one captain is to be selected.
∴ Required number of ways = (⁴C₁ × ⁶C₃+ ⁶C₄) × ⁴C₁ = (80 + 15) × 4 = 380