Permutations & Combinations – II

Here we are given seven digits, no two of which are alike
i. Required number of 4-digit numbers formed out of given seven digits =⁷P₄ = 7*6*5*4 = 840
ii. A number is divisible by 4 if the number formed by the last two digits taken together is divisible by 4 and here, we have 10 such possible cases viz. 12, 16, 24, 32, 36, 52, 56, 64, 72, 76
Now the remaining two first places on the left of 4-digit numbers are to be filled from the remaining 5 digits and this can be done in ⁵P₂ ways i.e., 5*4 i.e. 20 ways
Therefore, by fundamental theorem of counting, the required number = 10*20 = 200
iii. A number is divisible by 25 if the last two digits taken together form a number which is divisible by 25 and here, we have only 2 such possible cases viz. 25, 75.
Now the remaining two first places on the left of 4-digit numbers are to be filled from the remaining five digits and this can be done in ⁵P₂2 i.e., 5*4 i.e. 20 ways. Therefore, by fundamental theorem, the required number = 2*20 = 40

Here we are given five digits, two of which are alike and three are different. Numbers greater than 50000 will begin with either 5 or 9 and will be of five digits. If 5 is fixed at the first place on the left (i.e., in the beginning) then we are left with four places to be filled with remaining four digits two of which are alike, and this can be done in 4!/2! i.e., (4*3*2*1)/(2*1) i.e., 12 ways.
Similarly, if 9 is fixed at the first place on the left, then the number of 5-digit numbers beginning with 9 which can be formed out of the given digits is 12.
Therefore, required number of the numbers greater than 50000 which can be formed out of the given digits = 12+12 = 24

Here we are given five digits which are all different. Therefore, required numbers will be of five digits and not begin with either 1 or 21 —– ❶
Now total number of 5-digit numbers which can be formed out of given 5-digits = 5! —– ❷
Total number of 5-digit numbers beginning with 1 = Number of ways in which remaining four places can be filled with four remaining digits, when 1 is fixed at the extreme left position = ⁴P₄ = 4! —– ❸
And total number of 5-digit numbers beginning with 21 = Number of ways in which remaining three places can be filled with remaining three digits when 5-digit numbers start with 21 = ³P₃ = 3! —– ❹
Therefore, from ❶, with the help of ❷, ❸ and ❹, the required number = 5!-(4!-3!) = 120 – (24+6) = 90

Here we are given six digits, out of which three are alike and three are different.
Now the numbers which are greater than 200000 will be of six digits and will not begin with 0 —– ❶
Total number of 6-digit numbers which can be formed out of the given six digits is = 6!/3! = 120 —– ❷
Total number of 6-digit numbers which begin with 0 (and are therefore not 6-digit numbers) = number of ways in which remaining five places can be filled with five remaining digits, when 0 is fixed at the extreme left position = 5!/3! , since 3 digits are alike = (5*4*3*2*1)/(3*2*1) = 20 —– ❸
Therefore, from ❶ with the help of ❷ and ❸, the required number
= 120-20 = 100

There are 9 different letters in the given word, so the letters of the given word can be arranged in 9! ways, which include the one way given in the problem. Hence the required number of other ways = 9! -1 = 362879

In the word TRIANGLE, there are eight different letters, which can be arranged in 8! ways. Total number of permutations = 8! = 40320
Now, when T is fixed at the first place and E at the last place, the remaining six places are to be filled with remaining six letters and these six letters can be arranged in 6! ways.
Therefore, the number of permutations which begin with T and end with E = 6! = 720

There are 8 letters in the given word, out of which 3 are consonants (viz. Q, T, N) and 5 vowels (viz. A, E, I, O, U).
Now we are to form words containing 8 letters, beginning and ending with consonants. Thus, two places viz. the first and the last, are to be occupied by 2 consonants out of 3 and this can be done in ³P₂ i.e. 3*2 i.e. 6 ways.
Between these two places there are six more places to be filled with remaining six letters (viz. 5 vowels and 1 consonant) and these six can be arranged in these six places in 6! ways.
Hence, the required number of words = 6*6! = 6*720 = 4320

The given word contains 8 letters out of which 4 (viz. A, E, I, O) are vowels and remaining 4 are consonants.
i. The 4 vowels can be arranged in 4 odd places in 4! ways and the 4 consonants can be arranged in the remaining 4 places in 4! ways.
Therefore, the required number of words = 4!*4! = 24*24 = 576
ii. When O is fixed in the first place (i.e. in beginning) and E is fixed in the last place (i.e. in end) then remaining six letters are to arranged at the six places left and this can be done in 6! ways. Therefore, in this case, the required number of words = 6! = 720

The given word contains 10 letters, out of which 4 are vowels and 6 are consonants. We are to form words without altering the relative positions of vowels and consonants i.e., vowel can occupy the position of another vowel and a consonant that of another consonant.
Now the given 4 vowels, out of which two A’s are alike, can be arranged in 4!/2! ways. Similarly, the given 6 consonants, out of which two P’s and two T’s are alike, can be arranged in 6!/(2!*2!) ways.
The required number of words = (4!/2!)*(6!/(2!*2!)) = 2160

We can make up to 6-digit number using the digits 0, 3 and 6
Total six-digit numbers formed = 2*3⁵ (because the numbers cannot start with a 0)
Similarly, total number of 5-digit numbers formed = 2*3⁴
Therefore, total such numbers are 2(3⁵+3⁴+……3⁰)
= 2[1(3⁶-1)/(3-1)] = 728