Question

A seven digits number with distinct digits is in form of abcdefg. (g,f,e etc are digits at units, tens, hundred place etc.) where a < b < c < d > e > f > g. The number of such numbers are:
(a) 1951
(b) 2009
(c) 1560
(d) 1870

Answer 1

Hint: To begin with, we will find the possible values for d. Then for each value, we will find the number of all possible numbers. It is to be kept in mind that the a cannot be 0, as it is given that a number is a seven-digit number.

Complete step-by-step solution:
It is given that we have to find all the 7 digit numbers abcdefg such that a < b < c < d > e > f > g. For selecting ‘r’ elements from ‘n’ elements we use the formula of combination as $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
The available digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. It is given that d is greater than all the other digits in the number. Thus, d can take the values of 6, 7, 8, and 9 only.
Let us consider that if d = 6.
Then possible values for all the other digits will be 0, 1, 2, 3, 4, and 5, other digits are less than d.
Now, we know that a cannot be 0. Thus, a, b, c can be any three of the remaining 5.
Therefore, the number of ways to choose a, b, c will be $^{5}{{C}_{3}}=10$.
The remaining 3 digits will occupy e, g, and f in ascending order.
Thus, the number of numbers with d = 6 is 10.
Now if d = 7.
a, b, c can be any three from 1, 2, 3, 4, 5, and 6, since d is 7 and a cannot be 0.
Thus, the number of ways of choosing a, b, c (3 digits) from (1, 2, 3, 4, 5, and 6) will be $^{6}{{C}_{3}}=20$.
At this point, we know a, b, c, and d. Next e, g, and f can be occupied by any three of the remaining 3 from (1, 2, 3, 4, 5 and 6) or 0. The number of ways of choosing e, g, and f from 4 digits will be $^{4}{{C}_{3}}=4$.
Therefore, number of numbers such that d = 7 is $20\times 4=80$.
Let d = 8
a, b, c can be any three from 1, 2, 3, 4, 5, 6, and 7 since d is 8 and a cannot be 0.
Thus, the number of ways of choosing a, b, c from the available 7 digits will be $^{7}{{C}_{3}}=35$.
At this point, we know a, b, c, and d. Next e, g, and f can be occupied by any three of the remaining 4 digits after a, b, and c are selected or the digit 0. The number of ways of choosing e, g, and f from 5 digits will be $^{5}{{C}_{3}}=10$.
Therefore, number of numbers such that d = 8 is $35\times 10=350$.
Similarly, if d = 9
a, b, c can be any three from 1, 2, 3, 4, 5, 6, 7, and 8.
Thus, the number of ways of choosing a, b, c will be $^{8}{{C}_{3}}=56$.
At this point, we know a, b, c, and d. Next e, g, and f can be occupied by any three of the remaining 5 digits or a 0. The number of ways of choosing e, g, and f from the 6 available digits will be $^{6}{{C}_{3}}=20$.
Therefore, number of numbers such that d = 8 is $56\times 20=1120$.
Hence, the total number of arrangements which meet our condition will be the sum of arrangements with d = 6, 7, 8, and 9.
So, total number of arrangements = $10 + 80 + 350 + 1120 = 1560$
Thus, number of numbers such that a < b < c < d > e > f > g is (10 + 80 + 350 + 1120) = 1560. Hence, option (c) is the correct option.

Note: The number of ways of choosing r items from n items is given as $^{n}{{C}_{r}}$, where $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We are using combination and not permutations because once the number is chosen, they can be arranged only in one specific order. Permutations involve both choosing the numbers and arranging them in all possible ways.